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2x^2+8x-13=0
a = 2; b = 8; c = -13;
Δ = b2-4ac
Δ = 82-4·2·(-13)
Δ = 168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{42}}{2*2}=\frac{-8-2\sqrt{42}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{42}}{2*2}=\frac{-8+2\sqrt{42}}{4} $
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