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4x^2+48x-120=0
a = 4; b = 48; c = -120;
Δ = b2-4ac
Δ = 482-4·4·(-120)
Δ = 4224
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4224}=\sqrt{64*66}=\sqrt{64}*\sqrt{66}=8\sqrt{66}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{66}}{2*4}=\frac{-48-8\sqrt{66}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{66}}{2*4}=\frac{-48+8\sqrt{66}}{8} $
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