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100x-3x^2-15=0
a = -3; b = 100; c = -15;
Δ = b2-4ac
Δ = 1002-4·(-3)·(-15)
Δ = 9820
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9820}=\sqrt{4*2455}=\sqrt{4}*\sqrt{2455}=2\sqrt{2455}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-2\sqrt{2455}}{2*-3}=\frac{-100-2\sqrt{2455}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+2\sqrt{2455}}{2*-3}=\frac{-100+2\sqrt{2455}}{-6} $
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