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8x^2-3x-26=0
a = 8; b = -3; c = -26;
Δ = b2-4ac
Δ = -32-4·8·(-26)
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-29}{2*8}=\frac{-26}{16} =-1+5/8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+29}{2*8}=\frac{32}{16} =2 $
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