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6x^2-23x-35=0
a = 6; b = -23; c = -35;
Δ = b2-4ac
Δ = -232-4·6·(-35)
Δ = 1369
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{1369}=37$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-37}{2*6}=\frac{-14}{12} =-1+1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+37}{2*6}=\frac{60}{12} =5 $
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