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140=16x^2+32x
We move all terms to the left:
140-(16x^2+32x)=0
We get rid of parentheses
-16x^2-32x+140=0
a = -16; b = -32; c = +140;
Δ = b2-4ac
Δ = -322-4·(-16)·140
Δ = 9984
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{9984}=\sqrt{256*39}=\sqrt{256}*\sqrt{39}=16\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{39}}{2*-16}=\frac{32-16\sqrt{39}}{-32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{39}}{2*-16}=\frac{32+16\sqrt{39}}{-32} $
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