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x(x-25)=1350
We move all terms to the left:
x(x-25)-(1350)=0
We multiply parentheses
x^2-25x-1350=0
a = 1; b = -25; c = -1350;
Δ = b2-4ac
Δ = -252-4·1·(-1350)
Δ = 6025
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6025}=\sqrt{25*241}=\sqrt{25}*\sqrt{241}=5\sqrt{241}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{241}}{2*1}=\frac{25-5\sqrt{241}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{241}}{2*1}=\frac{25+5\sqrt{241}}{2} $
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