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32x^2+28x-3=0
a = 32; b = 28; c = -3;
Δ = b2-4ac
Δ = 282-4·32·(-3)
Δ = 1168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{73}}{2*32}=\frac{-28-4\sqrt{73}}{64} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{73}}{2*32}=\frac{-28+4\sqrt{73}}{64} $
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