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4x^2+40x-15=0
a = 4; b = 40; c = -15;
Δ = b2-4ac
Δ = 402-4·4·(-15)
Δ = 1840
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1840}=\sqrt{16*115}=\sqrt{16}*\sqrt{115}=4\sqrt{115}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{115}}{2*4}=\frac{-40-4\sqrt{115}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{115}}{2*4}=\frac{-40+4\sqrt{115}}{8} $
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