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4x^2+48x-128=0
a = 4; b = 48; c = -128;
Δ = b2-4ac
Δ = 482-4·4·(-128)
Δ = 4352
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4352}=\sqrt{256*17}=\sqrt{256}*\sqrt{17}=16\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-16\sqrt{17}}{2*4}=\frac{-48-16\sqrt{17}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+16\sqrt{17}}{2*4}=\frac{-48+16\sqrt{17}}{8} $
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