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x^2+40x-525=0
a = 1; b = 40; c = -525;
Δ = b2-4ac
Δ = 402-4·1·(-525)
Δ = 3700
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3700}=\sqrt{100*37}=\sqrt{100}*\sqrt{37}=10\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{37}}{2*1}=\frac{-40-10\sqrt{37}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{37}}{2*1}=\frac{-40+10\sqrt{37}}{2} $
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