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5x^2+10x=625
We move all terms to the left:
5x^2+10x-(625)=0
a = 5; b = 10; c = -625;
Δ = b2-4ac
Δ = 102-4·5·(-625)
Δ = 12600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12600}=\sqrt{900*14}=\sqrt{900}*\sqrt{14}=30\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-30\sqrt{14}}{2*5}=\frac{-10-30\sqrt{14}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+30\sqrt{14}}{2*5}=\frac{-10+30\sqrt{14}}{10} $
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