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10x^2+5x-100=0
a = 10; b = 5; c = -100;
Δ = b2-4ac
Δ = 52-4·10·(-100)
Δ = 4025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4025}=\sqrt{25*161}=\sqrt{25}*\sqrt{161}=5\sqrt{161}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{161}}{2*10}=\frac{-5-5\sqrt{161}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{161}}{2*10}=\frac{-5+5\sqrt{161}}{20} $
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