a^2=-4a+25

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Solution for a^2=-4a+25 equation:



a^2=-4a+25
We move all terms to the left:
a^2-(-4a+25)=0
We get rid of parentheses
a^2+4a-25=0
a = 1; b = 4; c = -25;
Δ = b2-4ac
Δ = 42-4·1·(-25)
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{29}}{2*1}=\frac{-4-2\sqrt{29}}{2} $
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{29}}{2*1}=\frac{-4+2\sqrt{29}}{2} $

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