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(2a-4)(a+3)=0
We multiply parentheses ..
(+2a^2+6a-4a-12)=0
We get rid of parentheses
2a^2+6a-4a-12=0
We add all the numbers together, and all the variables
2a^2+2a-12=0
a = 2; b = 2; c = -12;
Δ = b2-4ac
Δ = 22-4·2·(-12)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*2}=\frac{-12}{4} =-3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*2}=\frac{8}{4} =2 $
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