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8x^2+2x=10
We move all terms to the left:
8x^2+2x-(10)=0
a = 8; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·8·(-10)
Δ = 324
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*8}=\frac{-20}{16} =-1+1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*8}=\frac{16}{16} =1 $
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