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2x^2=12x+24
We move all terms to the left:
2x^2-(12x+24)=0
We get rid of parentheses
2x^2-12x-24=0
a = 2; b = -12; c = -24;
Δ = b2-4ac
Δ = -122-4·2·(-24)
Δ = 336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{21}}{2*2}=\frac{12-4\sqrt{21}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{21}}{2*2}=\frac{12+4\sqrt{21}}{4} $
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