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4x^2+120x-252=0
a = 4; b = 120; c = -252;
Δ = b2-4ac
Δ = 1202-4·4·(-252)
Δ = 18432
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{18432}=\sqrt{9216*2}=\sqrt{9216}*\sqrt{2}=96\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-96\sqrt{2}}{2*4}=\frac{-120-96\sqrt{2}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+96\sqrt{2}}{2*4}=\frac{-120+96\sqrt{2}}{8} $
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