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162x^2-15x-128=0
a = 162; b = -15; c = -128;
Δ = b2-4ac
Δ = -152-4·162·(-128)
Δ = 83169
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{83169}=\sqrt{9*9241}=\sqrt{9}*\sqrt{9241}=3\sqrt{9241}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{9241}}{2*162}=\frac{15-3\sqrt{9241}}{324} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{9241}}{2*162}=\frac{15+3\sqrt{9241}}{324} $
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