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5(x^2-3x)=625
We move all terms to the left:
5(x^2-3x)-(625)=0
We multiply parentheses
5x^2-15x-625=0
a = 5; b = -15; c = -625;
Δ = b2-4ac
Δ = -152-4·5·(-625)
Δ = 12725
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12725}=\sqrt{25*509}=\sqrt{25}*\sqrt{509}=5\sqrt{509}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5\sqrt{509}}{2*5}=\frac{15-5\sqrt{509}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{509}}{2*5}=\frac{15+5\sqrt{509}}{10} $
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