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2x^2+8x-252=0
a = 2; b = 8; c = -252;
Δ = b2-4ac
Δ = 82-4·2·(-252)
Δ = 2080
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2080}=\sqrt{16*130}=\sqrt{16}*\sqrt{130}=4\sqrt{130}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{130}}{2*2}=\frac{-8-4\sqrt{130}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{130}}{2*2}=\frac{-8+4\sqrt{130}}{4} $
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