Solve x^2+25>10x

$$x ^ { 2 } + 25 > 10 x$$

Solve for x

$$x\neq 5$$

Steps Using the Quadratic Formula

Subtract 10x from both sides.

$$x^{2}+25-10x>0$$

To solve the inequality, factor the left hand side. 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.

$$x^{2}+25-10x=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -10 for b, and 25 for c in the quadratic formula.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 1\times 25}}{2}$$

Do the calculations.

$$x=\frac{10±0}{2}$$

Solutions are the same.

$$x=5$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-5\right)^{2}>0$$

Inequality holds for x\neq 5.

$$x\neq 5$$