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4x^2+144x-95=0
a = 4; b = 144; c = -95;
Δ = b2-4ac
Δ = 1442-4·4·(-95)
Δ = 22256
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{22256}=\sqrt{16*1391}=\sqrt{16}*\sqrt{1391}=4\sqrt{1391}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(144)-4\sqrt{1391}}{2*4}=\frac{-144-4\sqrt{1391}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(144)+4\sqrt{1391}}{2*4}=\frac{-144+4\sqrt{1391}}{8} $
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