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16x^2-3=40x
We move all terms to the left:
16x^2-3-(40x)=0
a = 16; b = -40; c = -3;
Δ = b2-4ac
Δ = -402-4·16·(-3)
Δ = 1792
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16\sqrt{7}}{2*16}=\frac{40-16\sqrt{7}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16\sqrt{7}}{2*16}=\frac{40+16\sqrt{7}}{32} $
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