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3x(2x-5)=25+4x
We move all terms to the left:
3x(2x-5)-(25+4x)=0
We add all the numbers together, and all the variables
3x(2x-5)-(4x+25)=0
We multiply parentheses
6x^2-15x-(4x+25)=0
We get rid of parentheses
6x^2-15x-4x-25=0
We add all the numbers together, and all the variables
6x^2-19x-25=0
a = 6; b = -19; c = -25;
Δ = b2-4ac
Δ = -192-4·6·(-25)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-31}{2*6}=\frac{-12}{12} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+31}{2*6}=\frac{50}{12} =4+1/6 $
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