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12x^2+8x=158
We move all terms to the left:
12x^2+8x-(158)=0
a = 12; b = 8; c = -158;
Δ = b2-4ac
Δ = 82-4·12·(-158)
Δ = 7648
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7648}=\sqrt{16*478}=\sqrt{16}*\sqrt{478}=4\sqrt{478}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{478}}{2*12}=\frac{-8-4\sqrt{478}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{478}}{2*12}=\frac{-8+4\sqrt{478}}{24} $
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