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4x^2+20x-135=0
a = 4; b = 20; c = -135;
Δ = b2-4ac
Δ = 202-4·4·(-135)
Δ = 2560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2560}=\sqrt{256*10}=\sqrt{256}*\sqrt{10}=16\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{10}}{2*4}=\frac{-20-16\sqrt{10}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{10}}{2*4}=\frac{-20+16\sqrt{10}}{8} $
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