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a^2-15a+25=0
a = 1; b = -15; c = +25;
Δ = b2-4ac
Δ = -152-4·1·25
Δ = 125
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5\sqrt{5}}{2*1}=\frac{15-5\sqrt{5}}{2} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{5}}{2*1}=\frac{15+5\sqrt{5}}{2} $
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