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0=4x^2-5x-6-4x+3
We move all terms to the left:
0-(4x^2-5x-6-4x+3)=0
We add all the numbers together, and all the variables
-(4x^2-5x-6-4x+3)=0
We get rid of parentheses
-4x^2+5x+4x+6-3=0
We add all the numbers together, and all the variables
-4x^2+9x+3=0
a = -4; b = 9; c = +3;
Δ = b2-4ac
Δ = 92-4·(-4)·3
Δ = 129
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{129}}{2*-4}=\frac{-9-\sqrt{129}}{-8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{129}}{2*-4}=\frac{-9+\sqrt{129}}{-8} $
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