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4x^2-32x+7=5
We move all terms to the left:
4x^2-32x+7-(5)=0
We add all the numbers together, and all the variables
4x^2-32x+2=0
a = 4; b = -32; c = +2;
Δ = b2-4ac
Δ = -322-4·4·2
Δ = 992
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{992}=\sqrt{16*62}=\sqrt{16}*\sqrt{62}=4\sqrt{62}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{62}}{2*4}=\frac{32-4\sqrt{62}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{62}}{2*4}=\frac{32+4\sqrt{62}}{8} $
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