2q^2-3q=180

Solution for 2q^2-3q=180 equation:

2q^2-3q=180

We move all terms to the left:

2q^2-3q-(180)=0

a = 2; b = -3; c = -180;
Δ = b2-4ac
Δ = -32-4·2·(-180)
Δ = 1449
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1449}=\sqrt{9*161}=\sqrt{9}*\sqrt{161}=3\sqrt{161}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{161}}{2*2}=\frac{3-3\sqrt{161}}{4}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{161}}{2*2}=\frac{3+3\sqrt{161}}{4}
$