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0=6x^2-29x-5
We move all terms to the left:
0-(6x^2-29x-5)=0
We add all the numbers together, and all the variables
-(6x^2-29x-5)=0
We get rid of parentheses
-6x^2+29x+5=0
a = -6; b = 29; c = +5;
Δ = b2-4ac
Δ = 292-4·(-6)·5
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-31}{2*-6}=\frac{-60}{-12} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+31}{2*-6}=\frac{2}{-12} =-1/6 $
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