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3x^2-40x-975=0
a = 3; b = -40; c = -975;
Δ = b2-4ac
Δ = -402-4·3·(-975)
Δ = 13300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{13300}=\sqrt{100*133}=\sqrt{100}*\sqrt{133}=10\sqrt{133}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-10\sqrt{133}}{2*3}=\frac{40-10\sqrt{133}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+10\sqrt{133}}{2*3}=\frac{40+10\sqrt{133}}{6} $
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