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6x^2=9x+10
We move all terms to the left:
6x^2-(9x+10)=0
We get rid of parentheses
6x^2-9x-10=0
a = 6; b = -9; c = -10;
Δ = b2-4ac
Δ = -92-4·6·(-10)
Δ = 321
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{-(-9)-\sqrt{321}}{2*6}=\frac{9-\sqrt{321}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{321}}{2*6}=\frac{9+\sqrt{321}}{12} $
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