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(2a-6)(2a-6)+(a+2)=15
We move all terms to the left:
(2a-6)(2a-6)+(a+2)-(15)=0
We get rid of parentheses
(2a-6)(2a-6)+a+2-15=0
We multiply parentheses ..
(+4a^2-12a-12a+36)+a+2-15=0
We add all the numbers together, and all the variables
(+4a^2-12a-12a+36)+a-13=0
We get rid of parentheses
4a^2-12a-12a+a+36-13=0
We add all the numbers together, and all the variables
4a^2-23a+23=0
a = 4; b = -23; c = +23;
Δ = b2-4ac
Δ = -232-4·4·23
Δ = 161
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{161}}{2*4}=\frac{23-\sqrt{161}}{8} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{161}}{2*4}=\frac{23+\sqrt{161}}{8} $
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