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-x^{2}+10x+28=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\times 28}}{2\left(-1\right)}
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{-10±\sqrt{100-4\left(-1\right)\times 28}}{2\left(-1\right)}
Square 10.
x=\frac{-10±\sqrt{100+4\times 28}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-10±\sqrt{100+112}}{2\left(-1\right)}
Multiply 4 times 28.
x=\frac{-10±\sqrt{212}}{2\left(-1\right)}
Add 100 to 112.
x=\frac{-10±2\sqrt{53}}{2\left(-1\right)}
Take the square root of 212.
x=\frac{-10±2\sqrt{53}}{-2}
Multiply 2 times -1.
x=\frac{2\sqrt{53}-10}{-2}
Now solve the equation x=\frac{-10±2\sqrt{53}}{-2} when ± is plus. Add -10 to 2\sqrt{53}.
x=5-\sqrt{53}
Divide -10+2\sqrt{53} by -2.
x=\frac{-2\sqrt{53}-10}{-2}
Now solve the equation x=\frac{-10±2\sqrt{53}}{-2} when ± is minus. Subtract 2\sqrt{53} from -10.
x=\sqrt{53}+5
Divide -10-2\sqrt{53} by -2.
-x^{2}+10x+28=-\left(x-\left(5-\sqrt{53}\right)\right)\left(x-\left(\sqrt{53}+5\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5-\sqrt{53} for x_{1} and 5+\sqrt{53} for x_{2}.