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a^2+8a-13=0
a = 1; b = 8; c = -13;
Δ = b2-4ac
Δ = 82-4·1·(-13)
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{29}}{2*1}=\frac{-8-2\sqrt{29}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{29}}{2*1}=\frac{-8+2\sqrt{29}}{2} $
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