Solve 55x+5x^2=

$$55 x + 5 x ^ { 2 } =$$

Factor

$$5x\left(x+11\right)$$

Solution Steps

Factor out 5.

$$5\left(11x+x^{2}\right)$$

Consider 11x+x^{2}. Factor out x.

$$x\left(11+x\right)$$

Rewrite the complete factored expression.

$$5x\left(x+11\right)$$

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.

$$5x^{2}+55x=0$$

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.

$$x=\frac{-55±\sqrt{55^{2}}}{2\times 5}$$

Take the square root of 55^{2}.

$$x=\frac{-55±55}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{-55±55}{10}$$

Now solve the equation x=\frac{-55±55}{10} when ± is plus. Add -55 to 55.

$$x=\frac{0}{10}$$

Divide 0 by 10.

$$x=0$$

Now solve the equation x=\frac{-55±55}{10} when ± is minus. Subtract 55 from -55.

$$x=-\frac{110}{10}$$

Divide -110 by 10.

$$x=-11$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 0 for x_{1} and -11 for x_{2}.

$$5x^{2}+55x=5x\left(x-\left(-11\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$5x^{2}+55x=5x\left(x+11\right)$$

Evaluate

$$5x\left(x+11\right)$$