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4x^2+16x-600=0
a = 4; b = 16; c = -600;
Δ = b2-4ac
Δ = 162-4·4·(-600)
Δ = 9856
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9856}=\sqrt{64*154}=\sqrt{64}*\sqrt{154}=8\sqrt{154}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{154}}{2*4}=\frac{-16-8\sqrt{154}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{154}}{2*4}=\frac{-16+8\sqrt{154}}{8} $
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