50=(2x^2)+5x

Solution for 50=(2x^2)+5x equation:

50=(2x^2)+5x

We move all terms to the left:

50-((2x^2)+5x)=0

We get rid of parentheses

-2x^2-5x+50=0

a = -2; b = -5; c = +50;
Δ = b2-4ac
Δ = -52-4·(-2)·50
Δ = 425
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5\sqrt{17}}{2*-2}=\frac{5-5\sqrt{17}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5\sqrt{17}}{2*-2}=\frac{5+5\sqrt{17}}{-4}
$