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x^{2}+150x+5609=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.
x=\frac{-150±\sqrt{150^{2}-4\times 5609}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 150 for b, and 5609 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-150±\sqrt{22500-4\times 5609}}{2}
Square 150.
x=\frac{-150±\sqrt{22500-22436}}{2}
Multiply -4 times 5609.
x=\frac{-150±\sqrt{64}}{2}
Add 22500 to -22436.
x=\frac{-150±8}{2}
Take the square root of 64.
x=-\frac{142}{2}
Now solve the equation x=\frac{-150±8}{2} when ± is plus. Add -150 to 8.
x=-71
Divide -142 by 2.
x=-\frac{158}{2}
Now solve the equation x=\frac{-150±8}{2} when ± is minus. Subtract 8 from -150.
x=-79
Divide -158 by 2.
x=-71 x=-79
The equation is now solved.
x^{2}+150x+5609=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
x^{2}+150x+5609-5609=-5609
Subtract 5609 from both sides of the equation.
x^{2}+150x=-5609
Subtracting 5609 from itself leaves 0.
x^{2}+150x+75^{2}=-5609+75^{2}
Divide 150, the coefficient of the x term, by 2 to get 75. Then add the square of 75 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+150x+5625=-5609+5625
Square 75.
x^{2}+150x+5625=16
Add -5609 to 5625.
\left(x+75\right)^{2}=16
Factor x^{2}+150x+5625. 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(x+75\right)^{2}}=\sqrt{16}
Take the square root of both sides of the equation.
x+75=4 x+75=-4
Simplify.
x=-71 x=-79
Subtract 75 from both sides of the equation.