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5x^2+2x=25
We move all terms to the left:
5x^2+2x-(25)=0
a = 5; b = 2; c = -25;
Δ = b2-4ac
Δ = 22-4·5·(-25)
Δ = 504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{14}}{2*5}=\frac{-2-6\sqrt{14}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{14}}{2*5}=\frac{-2+6\sqrt{14}}{10} $
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