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(3x-2)(4x+3)=0
We multiply parentheses ..
(+12x^2+9x-8x-6)=0
We get rid of parentheses
12x^2+9x-8x-6=0
We add all the numbers together, and all the variables
12x^2+x-6=0
a = 12; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·12·(-6)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*12}=\frac{-18}{24} =-3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*12}=\frac{16}{24} =2/3 $
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