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8x^2-5=-20x
We move all terms to the left:
8x^2-5-(-20x)=0
We get rid of parentheses
8x^2+20x-5=0
a = 8; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·8·(-5)
Δ = 560
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{560}=\sqrt{16*35}=\sqrt{16}*\sqrt{35}=4\sqrt{35}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{35}}{2*8}=\frac{-20-4\sqrt{35}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{35}}{2*8}=\frac{-20+4\sqrt{35}}{16} $
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