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2x^2=4x+12
We move all terms to the left:
2x^2-(4x+12)=0
We get rid of parentheses
2x^2-4x-12=0
a = 2; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·2·(-12)
Δ = 112
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{7}}{2*2}=\frac{4-4\sqrt{7}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{7}}{2*2}=\frac{4+4\sqrt{7}}{4} $
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