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3x(2x-5)-4x=6x+1
We move all terms to the left:
3x(2x-5)-4x-(6x+1)=0
We add all the numbers together, and all the variables
-4x+3x(2x-5)-(6x+1)=0
We multiply parentheses
6x^2-4x-15x-(6x+1)=0
We get rid of parentheses
6x^2-4x-15x-6x-1=0
We add all the numbers together, and all the variables
6x^2-25x-1=0
a = 6; b = -25; c = -1;
Δ = b2-4ac
Δ = -252-4·6·(-1)
Δ = 649
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{-(-25)-\sqrt{649}}{2*6}=\frac{25-\sqrt{649}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{649}}{2*6}=\frac{25+\sqrt{649}}{12} $
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