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X(X+12)=48
We move all terms to the left:
X(X+12)-(48)=0
We multiply parentheses
X^2+12X-48=0
a = 1; b = 12; c = -48;
Δ = b2-4ac
Δ = 122-4·1·(-48)
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*1}=\frac{-12-4\sqrt{21}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{21}}{2*1}=\frac{-12+4\sqrt{21}}{2} $
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