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6x^2+5x=40
We move all terms to the left:
6x^2+5x-(40)=0
a = 6; b = 5; c = -40;
Δ = b2-4ac
Δ = 52-4·6·(-40)
Δ = 985
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{-(5)-\sqrt{985}}{2*6}=\frac{-5-\sqrt{985}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{985}}{2*6}=\frac{-5+\sqrt{985}}{12} $
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