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(2x-12)(x+1)=0
We multiply parentheses ..
(+2x^2+2x-12x-12)=0
We get rid of parentheses
2x^2+2x-12x-12=0
We add all the numbers together, and all the variables
2x^2-10x-12=0
a = 2; b = -10; c = -12;
Δ = b2-4ac
Δ = -102-4·2·(-12)
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*2}=\frac{-4}{4} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*2}=\frac{24}{4} =6 $
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