Solve 18x^2+82x+98

$$18 { x }^{ 2 } +82x+98$$

Factor

$$2\left(9x^{2}+41x+49\right)$$

Solution Steps

Factor out 2. Polynomial 9x^{2}+41x+49 is not factored since it does not have any rational roots.

$$2\left(9x^{2}+41x+49\right)$$

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.

$$18x^{2}+82x+98=0$$

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.

$$x=\frac{-82±\sqrt{82^{2}-4\times 18\times 98}}{2\times 18}$$

Square 82.

$$x=\frac{-82±\sqrt{6724-4\times 18\times 98}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-82±\sqrt{6724-72\times 98}}{2\times 18}$$

Multiply -72 times 98.

$$x=\frac{-82±\sqrt{6724-7056}}{2\times 18}$$

Add 6724 to -7056.

$$x=\frac{-82±\sqrt{-332}}{2\times 18}$$

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

$$18x^{2}+82x+98$$

Evaluate

$$18x^{2}+82x+98$$