Solve 7x^2-28x+35

$$7 { x }^{ 2 } -28x+35$$

Factor

$$7\left(x^{2}-4x+5\right)$$

Solution Steps

Factor out 7. Polynomial x^{2}-4x+5 is not factored since it does not have any rational roots.

$$7\left(x^{2}-4x+5\right)$$

Steps Using the Quadratic Formula

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.

$$7x^{2}-28x+35=0$$

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.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 7\times 35}}{2\times 7}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 7\times 35}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-28\right)±\sqrt{784-28\times 35}}{2\times 7}$$

Multiply -28 times 35.

$$x=\frac{-\left(-28\right)±\sqrt{784-980}}{2\times 7}$$

Add 784 to -980.

$$x=\frac{-\left(-28\right)±\sqrt{-196}}{2\times 7}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$7x^{2}-28x+35$$

Evaluate

$$7\left(x^{2}-4x+5\right)$$