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4x^2+56x-10=0
a = 4; b = 56; c = -10;
Δ = b2-4ac
Δ = 562-4·4·(-10)
Δ = 3296
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3296}=\sqrt{16*206}=\sqrt{16}*\sqrt{206}=4\sqrt{206}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-4\sqrt{206}}{2*4}=\frac{-56-4\sqrt{206}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+4\sqrt{206}}{2*4}=\frac{-56+4\sqrt{206}}{8} $
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