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5x+25=5x^2
We move all terms to the left:
5x+25-(5x^2)=0
determiningTheFunctionDomain -5x^2+5x+25=0
a = -5; b = 5; c = +25;
Δ = b2-4ac
Δ = 52-4·(-5)·25
Δ = 525
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{525}=\sqrt{25*21}=\sqrt{25}*\sqrt{21}=5\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{21}}{2*-5}=\frac{-5-5\sqrt{21}}{-10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{21}}{2*-5}=\frac{-5+5\sqrt{21}}{-10} $
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