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8x^2+9x=10
We move all terms to the left:
8x^2+9x-(10)=0
a = 8; b = 9; c = -10;
Δ = b2-4ac
Δ = 92-4·8·(-10)
Δ = 401
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{-(9)-\sqrt{401}}{2*8}=\frac{-9-\sqrt{401}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{401}}{2*8}=\frac{-9+\sqrt{401}}{16} $
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