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20x^2+48x+24=0
a = 20; b = 48; c = +24;
Δ = b2-4ac
Δ = 482-4·20·24
Δ = 384
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{384}=\sqrt{64*6}=\sqrt{64}*\sqrt{6}=8\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{6}}{2*20}=\frac{-48-8\sqrt{6}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{6}}{2*20}=\frac{-48+8\sqrt{6}}{40} $
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