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5x^2-20x-28=0
a = 5; b = -20; c = -28;
Δ = b2-4ac
Δ = -202-4·5·(-28)
Δ = 960
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{960}=\sqrt{64*15}=\sqrt{64}*\sqrt{15}=8\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8\sqrt{15}}{2*5}=\frac{20-8\sqrt{15}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8\sqrt{15}}{2*5}=\frac{20+8\sqrt{15}}{10} $
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