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-4x^2+12x+24=0
a = -4; b = 12; c = +24;
Δ = b2-4ac
Δ = 122-4·(-4)·24
Δ = 528
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{33}}{2*-4}=\frac{-12-4\sqrt{33}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{33}}{2*-4}=\frac{-12+4\sqrt{33}}{-8} $
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