How do you solve \[2{{x}^{2}}-15x=8?\]

Hint: We are given \[2{{x}^{2}}-15x=8\] and to solve this we learn about the type of equations we are given and then learn the number of the solutions of equations. We will learn how to factor the quadratic equation, we will use the middle term split to factor the term and we will simplify by taking common terms out. We also use the zero product rule to get our answer. At the last, we will also learn about the quadratic formula for solving such equations in an easy way and more speedy way.

Complete step-by-step solution:
We are given \[2{{x}^{2}}-15x=8\] and we are asked to solve the given problem. First, we observe that it has a maximum power of ‘2’ so it is a quadratic equation. Now we should know that the quadratic equation has 2 solutions or we say an equation of power ‘n’ will have an ‘n’ solution. Now as it is a quadratic equation we will change it into standard form \[a{{x}^{2}}+bx+c=0.\] Now, subtracting 8 to both the sides, we get,
\[\Rightarrow 2{{x}^{2}}-15x-8=8-8\]
\[\Rightarrow 2{{x}^{2}}-15x-8=0\]
To solve this equation we will first take the greatest common factor possibly available to the terms. As we can see that in \[2{{x}^{2}}-15x-8=0,\] 2,15 and 8 has nothing in common so nothing can be taken out, so the equation remains the same. To solve the equation, we will use the middle term split. The middle term split method says for any quadratic equation.
1. We first product \[a\times c.\]
2. We try to find a number such that their sum or difference is ‘b’ while the product is the same as \[a\times c.\]
3. We use that and separate like the term.
4. We use the zero product rule to get our solution.
Now, we have \[2{{x}^{2}}-15x-8=0.\] So, we have a = 2, b = – 15 and c = – 8. So, \[a\times c=-16.\]
Now, we can see that \[-16\times 1=-16.\] And also, – 16 + 1 = – 15.
So, we use this to split the middle term ‘b’. So, \[2{{x}^{2}}-15x-18=0\] becomes
\[2{{x}^{2}}+\left( -16+1 \right)x-8=0\]
Now, opening the brackets, we get,
\[\Rightarrow 2{{x}^{2}}-16x+x-8=0\]
Now, taking common from the first two terms and the last two terms, we get,
\[\Rightarrow 2x\left( x-8 \right)+1\left( x-8 \right)=0\]
As x – 8 is common, so simplifying we get,
\[\Rightarrow \left( x-8 \right)\left( 2x+1 \right)=0\]
Now using zero product rules, we have
\[\Rightarrow x-8=0;2x+1=0\]
So, we get,
\[\Rightarrow x=8;x=\dfrac{-1}{2}\]
So, we get, x = 8 and \[x=\dfrac{-1}{2}\] as the solutions.

Note: Another way to solve this is we will use the quadratic formula. The quadratic formula states that for any \[a{{x}^{2}}+bx+c=0,\] the solution is given as \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.\] For \[2{{x}^{2}}-15x-8=0,\] we have a = 2, b = – 15 and c = – 8. So, using this we get,
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
\[\Rightarrow x=\dfrac{-\left( -15 \right)\pm \sqrt{{{\left( -15 \right)}^{2}}-4\times 2\times \left( -8 \right)}}{2\times 2}\]
On simplifying, we get,
\[\Rightarrow x=\dfrac{15\pm \sqrt{225+64}}{4}\]
So, we have
\[\Rightarrow x=\dfrac{15\pm \sqrt{289}}{4}\]
As \[\sqrt{289}=17,\] so we get,
\[\Rightarrow x=\dfrac{15\pm 17}{4}\]
So, the solution will be
\[x=\dfrac{15+17}{4};x=\dfrac{15-17}{4}\]
So, simplifying, we get,
\[\Rightarrow x=8;x=\dfrac{-1}{2}\]
Therefore the solutions are x = 8 and \[x=\dfrac{-1}{2}.\]