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s^{2}+16-8s=0
Subtract 8s from both sides.
s^{2}-8s+16=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-8 ab=16
To solve the equation, factor s^{2}-8s+16 using formula s^{2}+\left(a+b\right)s+ab=\left(s+a\right)\left(s+b\right). To find a and b, set up a system to be solved.
-1,-16 -2,-8 -4,-4
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.
-1-16=-17 -2-8=-10 -4-4=-8
Calculate the sum for each pair.
a=-4 b=-4
The solution is the pair that gives sum -8.
\left(s-4\right)\left(s-4\right)
Rewrite factored expression \left(s+a\right)\left(s+b\right) using the obtained values.
\left(s-4\right)^{2}
Rewrite as a binomial square.
s=4
To find equation solution, solve s-4=0.
s^{2}+16-8s=0
Subtract 8s from both sides.
s^{2}-8s+16=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=-8 ab=1\times 16=16
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as s^{2}+as+bs+16. To find a and b, set up a system to be solved.
-1,-16 -2,-8 -4,-4
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 16.
-1-16=-17 -2-8=-10 -4-4=-8
Calculate the sum for each pair.
a=-4 b=-4
The solution is the pair that gives sum -8.
\left(s^{2}-4s\right)+\left(-4s+16\right)
Rewrite s^{2}-8s+16 as \left(s^{2}-4s\right)+\left(-4s+16\right).
s\left(s-4\right)-4\left(s-4\right)
Factor out s in the first and -4 in the second group.
\left(s-4\right)\left(s-4\right)
Factor out common term s-4 by using distributive property.
\left(s-4\right)^{2}
Rewrite as a binomial square.
s=4
To find equation solution, solve s-4=0.
s^{2}+16-8s=0
Subtract 8s from both sides.
s^{2}-8s+16=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
s=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 16}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
s=\frac{-\left(-8\right)±\sqrt{64-4\times 16}}{2}
Square -8.
s=\frac{-\left(-8\right)±\sqrt{64-64}}{2}
Multiply -4 times 16.
s=\frac{-\left(-8\right)±\sqrt{0}}{2}
Add 64 to -64.
s=-\frac{-8}{2}
Take the square root of 0.
s=\frac{8}{2}
The opposite of -8 is 8.
s=4
Divide 8 by 2.
s^{2}+16-8s=0
Subtract 8s from both sides.
s^{2}-8s=-16
Subtract 16 from both sides. Anything subtracted from zero gives its negation.
s^{2}-8s+\left(-4\right)^{2}=-16+\left(-4\right)^{2}
Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
s^{2}-8s+16=-16+16
Square -4.
s^{2}-8s+16=0
Add -16 to 16.
\left(s-4\right)^{2}=0
Factor s^{2}-8s+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(s-4\right)^{2}}=\sqrt{0}
Take the square root of both sides of the equation.
s-4=0 s-4=0
Simplify.
s=4 s=4
Add 4 to both sides of the equation.
s=4
The equation is now solved. Solutions are the same.