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4x^2-16x+12=3x^2-x-14
We move all terms to the left:
4x^2-16x+12-(3x^2-x-14)=0
We get rid of parentheses
4x^2-3x^2-16x+x+14+12=0
We add all the numbers together, and all the variables
x^2-15x+26=0
a = 1; b = -15; c = +26;
Δ = b2-4ac
Δ = -152-4·1·26
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-11}{2*1}=\frac{4}{2} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+11}{2*1}=\frac{26}{2} =13 $
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