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6x+4x^2=29
We move all terms to the left:
6x+4x^2-(29)=0
a = 4; b = 6; c = -29;
Δ = b2-4ac
Δ = 62-4·4·(-29)
Δ = 500
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10\sqrt{5}}{2*4}=\frac{-6-10\sqrt{5}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10\sqrt{5}}{2*4}=\frac{-6+10\sqrt{5}}{8} $
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