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5x^2+28x-57=0
a = 5; b = 28; c = -57;
Δ = b2-4ac
Δ = 282-4·5·(-57)
Δ = 1924
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1924}=\sqrt{4*481}=\sqrt{4}*\sqrt{481}=2\sqrt{481}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-2\sqrt{481}}{2*5}=\frac{-28-2\sqrt{481}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+2\sqrt{481}}{2*5}=\frac{-28+2\sqrt{481}}{10} $
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