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4a^2-14a+8=0
a = 4; b = -14; c = +8;
Δ = b2-4ac
Δ = -142-4·4·8
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-2\sqrt{17}}{2*4}=\frac{14-2\sqrt{17}}{8} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+2\sqrt{17}}{2*4}=\frac{14+2\sqrt{17}}{8} $
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