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x^2+25x-225=0
a = 1; b = 25; c = -225;
Δ = b2-4ac
Δ = 252-4·1·(-225)
Δ = 1525
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1525}=\sqrt{25*61}=\sqrt{25}*\sqrt{61}=5\sqrt{61}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-5\sqrt{61}}{2*1}=\frac{-25-5\sqrt{61}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+5\sqrt{61}}{2*1}=\frac{-25+5\sqrt{61}}{2} $
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