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(2x)(3x+3)=180
We move all terms to the left:
(2x)(3x+3)-(180)=0
We multiply parentheses
6x^2+6x-180=0
a = 6; b = 6; c = -180;
Δ = b2-4ac
Δ = 62-4·6·(-180)
Δ = 4356
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4356}=66$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-66}{2*6}=\frac{-72}{12} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+66}{2*6}=\frac{60}{12} =5 $
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