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4x^2+15x-450=0
a = 4; b = 15; c = -450;
Δ = b2-4ac
Δ = 152-4·4·(-450)
Δ = 7425
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{7425}=\sqrt{225*33}=\sqrt{225}*\sqrt{33}=15\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15\sqrt{33}}{2*4}=\frac{-15-15\sqrt{33}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15\sqrt{33}}{2*4}=\frac{-15+15\sqrt{33}}{8} $
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