How do you solve $2{{x}^{2}}+5x-3=0$ ?

Hint: In the above question, we were asked to solve $2{{x}^{2}}+5x-3=0$. We will use middle term factorization to solve this problem. Also, we will get 2 solutions for this question as the middle term always gives two solutions. So let us see how we can solve this problem.

Complete Step by Step Solution:
We have to solve $2{{x}^{2}}+5x-3=0$ . We will factorize the equation and then we will get the solutions.
On factoring the above equation by splitting them into the middle term we get,
 $\Rightarrow 2{{x}^{2}}+5x-3=0$
 $\Rightarrow 2{{x}^{2}}+6x-x-3=0$
After taking common we get,
 $\Rightarrow 2x(x+3)-1(x+3)=0$
 $\Rightarrow (2x-1)(x+3)=0$
On equating 2x - 1 with 0 we get,
 $\Rightarrow 2x-1=0$
 $\therefore x=\dfrac{1}{2}$
On equating x + 3 with 0 we get,
 $\Rightarrow x+3=0$
 $\therefore x=-3$

Therefore, on solving $2{{x}^{2}}+5x-3=0$ we get $x=\dfrac{1}{2}$ or x = - 3

Note:
The above problem can also be solved with the quadratic formula. Let’s see how we can solve this problem with the quadratic formula.
 $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is the quadratic formula which we will use to solve the problem
 $2{{x}^{2}}+5x-3=0$
From this equation we can identify that
a = 2
b = 5
c = -3
 $\therefore x=\dfrac{-5\pm \sqrt{{{5}^{2}}-4.2(-3)}}{2.2}$
Here also we will get 2 values of x because of the $\pm$ sign.
If we take the plus sign we will get
 $\Rightarrow x=\dfrac{-5+\sqrt{{{5}^{2}}-4.2(-3)}}{2.2}$
 $\Rightarrow x=\dfrac{-5+\sqrt{25+24}}{4}$
On solving this we will get,
 $\Rightarrow x=\dfrac{-5+\sqrt{49}}{4}$
 $\sqrt{49}$ is 7, so
 $\Rightarrow x=\dfrac{-5+7}{4}$
 $\Rightarrow x=\dfrac{2}{4}=\dfrac{1}{2}$
If we take the minus sign we will get,
 $\Rightarrow x=\dfrac{-5-\sqrt{{{5}^{2}}-4.2(-3)}}{2.2}$
 $\Rightarrow x=\dfrac{-5-\sqrt{25+24}}{4}$
 $\Rightarrow x=\dfrac{-5-\sqrt{49}}{4}$
$\sqrt{49}$ is 7
 $\Rightarrow x=\dfrac{-5-7}{4}$
 $\Rightarrow x=\dfrac{-12}{4}=-3$
Therefore, on solving $2{{x}^{2}}+5x-3=0$ we get $x=\dfrac{1}{2}$ or x = - 3
Also, both the formula gave us two solutions.