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(x)=8x^2+2x-5
We move all terms to the left:
(x)-(8x^2+2x-5)=0
We get rid of parentheses
-8x^2+x-2x+5=0
We add all the numbers together, and all the variables
-8x^2-1x+5=0
a = -8; b = -1; c = +5;
Δ = b2-4ac
Δ = -12-4·(-8)·5
Δ = 161
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{161}}{2*-8}=\frac{1-\sqrt{161}}{-16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{161}}{2*-8}=\frac{1+\sqrt{161}}{-16} $
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