Solve 5x^2-35x+x+7

$$5 x ^ { 2 } - 35 x + x + 7$$

Factor

$$5\left(x-\frac{17-\sqrt{254}}{5}\right)\left(x-\frac{\sqrt{254}+17}{5}\right)$$

Steps Using the Quadratic Formula

Combine -35x and x to get -34x.

$$factor(5x^{2}-34x+7)$$

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.

$$5x^{2}-34x+7=0$$

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.

$$x=\frac{-\left(-34\right)±\sqrt{\left(-34\right)^{2}-4\times 5\times 7}}{2\times 5}$$

Square -34.

$$x=\frac{-\left(-34\right)±\sqrt{1156-4\times 5\times 7}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-34\right)±\sqrt{1156-20\times 7}}{2\times 5}$$

Multiply -20 times 7.

$$x=\frac{-\left(-34\right)±\sqrt{1156-140}}{2\times 5}$$

Add 1156 to -140.

$$x=\frac{-\left(-34\right)±\sqrt{1016}}{2\times 5}$$

Take the square root of 1016.

$$x=\frac{-\left(-34\right)±2\sqrt{254}}{2\times 5}$$

The opposite of -34 is 34.

$$x=\frac{34±2\sqrt{254}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{34±2\sqrt{254}}{10}$$

Now solve the equation x=\frac{34±2\sqrt{254}}{10} when ± is plus. Add 34 to 2\sqrt{254}.

$$x=\frac{2\sqrt{254}+34}{10}$$

Divide 34+2\sqrt{254} by 10.

$$x=\frac{\sqrt{254}+17}{5}$$

Now solve the equation x=\frac{34±2\sqrt{254}}{10} when ± is minus. Subtract 2\sqrt{254} from 34.

$$x=\frac{34-2\sqrt{254}}{10}$$

Divide 34-2\sqrt{254} by 10.

$$x=\frac{17-\sqrt{254}}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{17+\sqrt{254}}{5} for x_{1} and \frac{17-\sqrt{254}}{5} for x_{2}.

$$5x^{2}-34x+7=5\left(x-\frac{\sqrt{254}+17}{5}\right)\left(x-\frac{17-\sqrt{254}}{5}\right)$$

Evaluate

$$5x^{2}-34x+7$$

Solution Steps

Combine -35x and x to get -34x.

$$5x^{2}-34x+7$$