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2x(3x+15)=36
We move all terms to the left:
2x(3x+15)-(36)=0
We multiply parentheses
6x^2+30x-36=0
a = 6; b = 30; c = -36;
Δ = b2-4ac
Δ = 302-4·6·(-36)
Δ = 1764
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{1764}=42$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-42}{2*6}=\frac{-72}{12} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+42}{2*6}=\frac{12}{12} =1 $
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