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5x^2+20x-5=0
a = 5; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·5·(-5)
Δ = 500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{5}}{2*5}=\frac{-20-10\sqrt{5}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{5}}{2*5}=\frac{-20+10\sqrt{5}}{10} $
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