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(2x+3)(3x-4)-3=0
We multiply parentheses ..
(+6x^2-8x+9x-12)-3=0
We get rid of parentheses
6x^2-8x+9x-12-3=0
We add all the numbers together, and all the variables
6x^2+x-15=0
a = 6; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·6·(-15)
Δ = 361
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{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-19}{2*6}=\frac{-20}{12} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+19}{2*6}=\frac{18}{12} =1+1/2 $
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