Solve 2x^2-15x+25x-5

$$2 x ^ { 2 } - 15 x + 25 x - 5$$

Factor

$$2\left(x-\frac{-\sqrt{35}-5}{2}\right)\left(x-\frac{\sqrt{35}-5}{2}\right)$$

Steps Using the Quadratic Formula

Combine -15x and 25x to get 10x.

$$factor(2x^{2}+10x-5)$$

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.

$$2x^{2}+10x-5=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\left(-5\right)}}{2\times 2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-10±\sqrt{100+40}}{2\times 2}$$

Add 100 to 40.

$$x=\frac{-10±\sqrt{140}}{2\times 2}$$

Take the square root of 140.

$$x=\frac{-10±2\sqrt{35}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-10±2\sqrt{35}}{4}$$

Now solve the equation x=\frac{-10±2\sqrt{35}}{4} when ± is plus. Add -10 to 2\sqrt{35}.

$$x=\frac{2\sqrt{35}-10}{4}$$

Divide -10+2\sqrt{35} by 4.

$$x=\frac{\sqrt{35}-5}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{35}}{4} when ± is minus. Subtract 2\sqrt{35} from -10.

$$x=\frac{-2\sqrt{35}-10}{4}$$

Divide -10-2\sqrt{35} by 4.

$$x=\frac{-\sqrt{35}-5}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5+\sqrt{35}}{2} for x_{1} and \frac{-5-\sqrt{35}}{2} for x_{2}.

$$2x^{2}+10x-5=2\left(x-\frac{\sqrt{35}-5}{2}\right)\left(x-\frac{-\sqrt{35}-5}{2}\right)$$

Evaluate

$$2x^{2}+10x-5$$

Solution Steps

Combine -15x and 25x to get 10x.

$$2x^{2}+10x-5$$