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(4x-3)(3x+4)=0
We multiply parentheses ..
(+12x^2+16x-9x-12)=0
We get rid of parentheses
12x^2+16x-9x-12=0
We add all the numbers together, and all the variables
12x^2+7x-12=0
a = 12; b = 7; c = -12;
Δ = b2-4ac
Δ = 72-4·12·(-12)
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-25}{2*12}=\frac{-32}{24} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+25}{2*12}=\frac{18}{24} =3/4 $
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