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X(3X+4)=40
We move all terms to the left:
X(3X+4)-(40)=0
We multiply parentheses
3X^2+4X-40=0
a = 3; b = 4; c = -40;
Δ = b2-4ac
Δ = 42-4·3·(-40)
Δ = 496
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{31}}{2*3}=\frac{-4-4\sqrt{31}}{6} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{31}}{2*3}=\frac{-4+4\sqrt{31}}{6} $
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