How do you solve \[2{{x}^{2}}-5x=4\]?

Hint:Firstly, we will find the discriminant of the given equation is \[D=57\] which is greater than zero. Therefore, the given equation has two real roots which can be found by using the quadratic formula, \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. So, we will get the roots by substituting a = 2, b = -5 and c = -4 in the formula.

Complete step by step solution:
According to the given question, we have to solve the above equation for \[x\] using the quadratic formula.
The given expression is, \[2{{x}^{2}}-5x-4=0\],
we will start with re-write the given equation into the standard form of a quadratic equation which is \[a{{x}^{2}}+bx+c=0\], we get,
\[2{{x}^{2}}-5x-4=0\]------(1)
Before applying the quadratic formula on the above expression, we will check what type of roots do the expression using discriminant, \[D\],
We know that
\[D={{b}^{2}}-4ac\]
On comparing the discriminant with the equation (1), we get the value of variables as,
\[a=2,b=-5,c=-4\]
On substituting these values in the discriminant formula, we get,
\[\Rightarrow D={{(-5)}^{2}}-4(2)(-4)\]
\[\Rightarrow D=25+32\]
\[\Rightarrow D=57>0\]
We have the value of \[D>0\], therefore the given equation has 2 real roots. And here we will use the quadratic formula to find the values of \[x\].
We have the quadratic formula as,
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Substituting the values of variables known, we get,
\[\Rightarrow x=\dfrac{-(-5)\pm \sqrt{(57)}}{2(2)}\]
\[\Rightarrow x=\dfrac{5\pm \sqrt{(57)}}{4}\]
\[\Rightarrow x=\dfrac{5+\sqrt{(57)}}{4},\dfrac{5-\sqrt{(57)}}{4}\]

Therefore, the two values of \[x=\dfrac{5+\sqrt{(57)}}{4},\dfrac{5-\sqrt{(57)}}{4}\].

Note: The discriminant used in the above solution is very important to know the number of values of \[x\] will have and also the type of solution the equation will have, that is, either real or imaginary. This also helps in verifying the values we get after solving the quadratic formula for the given equation.