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8x^2-28x+10=0
a = 8; b = -28; c = +10;
Δ = b2-4ac
Δ = -282-4·8·10
Δ = 464
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{29}}{2*8}=\frac{28-4\sqrt{29}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{29}}{2*8}=\frac{28+4\sqrt{29}}{16} $
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