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8x^2-3x=7
We move all terms to the left:
8x^2-3x-(7)=0
a = 8; b = -3; c = -7;
Δ = b2-4ac
Δ = -32-4·8·(-7)
Δ = 233
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{-(-3)-\sqrt{233}}{2*8}=\frac{3-\sqrt{233}}{16} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{233}}{2*8}=\frac{3+\sqrt{233}}{16} $
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