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5x^2+40x-11=0
a = 5; b = 40; c = -11;
Δ = b2-4ac
Δ = 402-4·5·(-11)
Δ = 1820
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1820}=\sqrt{4*455}=\sqrt{4}*\sqrt{455}=2\sqrt{455}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{455}}{2*5}=\frac{-40-2\sqrt{455}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{455}}{2*5}=\frac{-40+2\sqrt{455}}{10} $
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