Solve 6x^2-3x+30

$$6 { x }^{ 2 } -3x+30$$

Factor

$$3\left(2x^{2}-x+10\right)$$

Solution Steps

Factor out 3. Polynomial 2x^{2}-x+10 is not factored since it does not have any rational roots.

$$3\left(2x^{2}-x+10\right)$$

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.

$$6x^{2}-3x+30=0$$

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.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 6\times 30}}{2\times 6}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 6\times 30}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-3\right)±\sqrt{9-24\times 30}}{2\times 6}$$

Multiply -24 times 30.

$$x=\frac{-\left(-3\right)±\sqrt{9-720}}{2\times 6}$$

Add 9 to -720.

$$x=\frac{-\left(-3\right)±\sqrt{-711}}{2\times 6}$$

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

$$6x^{2}-3x+30$$

Evaluate

$$6x^{2}-3x+30$$