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4x^2+40x+90=0
a = 4; b = 40; c = +90;
Δ = b2-4ac
Δ = 402-4·4·90
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{10}}{2*4}=\frac{-40-4\sqrt{10}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{10}}{2*4}=\frac{-40+4\sqrt{10}}{8} $
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