(2x)^2+(3x)=50

Solution for (2x)^2+(3x)=50 equation:

(2x)^2+(3x)=50

We move all terms to the left:

(2x)^2+(3x)-(50)=0

a = 2; b = 3; c = -50;
Δ = b2-4ac
Δ = 32-4·2·(-50)
Δ = 409
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{409}}{2*2}=\frac{-3-\sqrt{409}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{409}}{2*2}=\frac{-3+\sqrt{409}}{4}
$