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4x^2+24x-140=0
a = 4; b = 24; c = -140;
Δ = b2-4ac
Δ = 242-4·4·(-140)
Δ = 2816
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2816}=\sqrt{256*11}=\sqrt{256}*\sqrt{11}=16\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16\sqrt{11}}{2*4}=\frac{-24-16\sqrt{11}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16\sqrt{11}}{2*4}=\frac{-24+16\sqrt{11}}{8} $
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