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x(3x+3)=78
We move all terms to the left:
x(3x+3)-(78)=0
We multiply parentheses
3x^2+3x-78=0
a = 3; b = 3; c = -78;
Δ = b2-4ac
Δ = 32-4·3·(-78)
Δ = 945
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{105}}{2*3}=\frac{-3-3\sqrt{105}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{105}}{2*3}=\frac{-3+3\sqrt{105}}{6} $
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