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7x^2+40x-36=0
a = 7; b = 40; c = -36;
Δ = b2-4ac
Δ = 402-4·7·(-36)
Δ = 2608
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2608}=\sqrt{16*163}=\sqrt{16}*\sqrt{163}=4\sqrt{163}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{163}}{2*7}=\frac{-40-4\sqrt{163}}{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{163}}{2*7}=\frac{-40+4\sqrt{163}}{14} $
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