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4g^2+8g-6=0
a = 4; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·4·(-6)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$g_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{10}}{2*4}=\frac{-8-4\sqrt{10}}{8} $$g_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{10}}{2*4}=\frac{-8+4\sqrt{10}}{8} $
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