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12x^2-35x+8=0
a = 12; b = -35; c = +8;
Δ = b2-4ac
Δ = -352-4·12·8
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-29}{2*12}=\frac{6}{24} =1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+29}{2*12}=\frac{64}{24} =2+2/3 $
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