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8x^2-3x-38=0
a = 8; b = -3; c = -38;
Δ = b2-4ac
Δ = -32-4·8·(-38)
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-35}{2*8}=\frac{-32}{16} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+35}{2*8}=\frac{38}{16} =2+3/8 $
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