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2x(3x+3)=76
We move all terms to the left:
2x(3x+3)-(76)=0
We multiply parentheses
6x^2+6x-76=0
a = 6; b = 6; c = -76;
Δ = b2-4ac
Δ = 62-4·6·(-76)
Δ = 1860
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1860}=\sqrt{4*465}=\sqrt{4}*\sqrt{465}=2\sqrt{465}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{465}}{2*6}=\frac{-6-2\sqrt{465}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{465}}{2*6}=\frac{-6+2\sqrt{465}}{12} $
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