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3x(2x+5)=36
We move all terms to the left:
3x(2x+5)-(36)=0
We multiply parentheses
6x^2+15x-36=0
a = 6; b = 15; c = -36;
Δ = b2-4ac
Δ = 152-4·6·(-36)
Δ = 1089
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{1089}=33$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-33}{2*6}=\frac{-48}{12} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+33}{2*6}=\frac{18}{12} =1+1/2 $
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