Solve 25-70x+49x^2geq0

$$25-70x+49 { x }^{ 2 } \geq 0$$

Solve for x

$$x\in \mathrm{R}$$

Steps Using the Quadratic Formula

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.

$$25-70x+49x^{2}=0$$

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 49 for a, -70 for b, and 25 for c in the quadratic formula.

$$x=\frac{-\left(-70\right)±\sqrt{\left(-70\right)^{2}-4\times 49\times 25}}{2\times 49}$$

Do the calculations.

$$x=\frac{70±0}{98}$$

Solutions are the same.

$$x=\frac{5}{7}$$

Rewrite the inequality by using the obtained solutions.

$$49\left(x-\frac{5}{7}\right)^{2}\geq 0$$

Inequality holds for x\in \mathrm{R}.

$$x\in \mathrm{R}$$