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10x^2-4x=8
We move all terms to the left:
10x^2-4x-(8)=0
a = 10; b = -4; c = -8;
Δ = b2-4ac
Δ = -42-4·10·(-8)
Δ = 336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{21}}{2*10}=\frac{4-4\sqrt{21}}{20} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{21}}{2*10}=\frac{4+4\sqrt{21}}{20} $
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