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(a+3)(2a-5)=5a-15
We move all terms to the left:
(a+3)(2a-5)-(5a-15)=0
We get rid of parentheses
(a+3)(2a-5)-5a+15=0
We multiply parentheses ..
(+2a^2-5a+6a-15)-5a+15=0
We get rid of parentheses
2a^2-5a+6a-5a-15+15=0
We add all the numbers together, and all the variables
2a^2-4a=0
a = 2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·2·0
Δ = 16
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{16}=4$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*2}=\frac{0}{4} =0 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*2}=\frac{8}{4} =2 $
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