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-4a^2+2a+3=0
a = -4; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-4)·3
Δ = 52
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*-4}=\frac{-2-2\sqrt{13}}{-8} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*-4}=\frac{-2+2\sqrt{13}}{-8} $
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