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4x^2+52x-240=0
a = 4; b = 52; c = -240;
Δ = b2-4ac
Δ = 522-4·4·(-240)
Δ = 6544
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6544}=\sqrt{16*409}=\sqrt{16}*\sqrt{409}=4\sqrt{409}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-4\sqrt{409}}{2*4}=\frac{-52-4\sqrt{409}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+4\sqrt{409}}{2*4}=\frac{-52+4\sqrt{409}}{8} $
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