(2x^2)+10x=36

Solution for (2x^2)+10x=36 equation:

(2x^2)+10x=36

We move all terms to the left:

(2x^2)+10x-(36)=0

a = 2; b = 10; c = -36;
Δ = b2-4ac
Δ = 102-4·2·(-36)
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{97}}{2*2}=\frac{-10-2\sqrt{97}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{97}}{2*2}=\frac{-10+2\sqrt{97}}{4}
$