Solve f(x)=-7x^2+3x

$$f ( x ) = - 7 x ^ { 2 } + 3 x$$

Factor

$$x\left(3-7x\right)$$

Solution Steps

Factor out x.

$$x\left(-7x+3\right)$$

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.

$$-7x^{2}+3x=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}}}{2\left(-7\right)}$$

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{-3±3}{-14}$$

Now solve the equation x=\frac{-3±3}{-14} when ± is plus. Add -3 to 3.

$$x=\frac{0}{-14}$$

Divide 0 by -14.

$$x=0$$

Now solve the equation x=\frac{-3±3}{-14} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{-14}$$

Reduce the fraction \frac{-6}{-14} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{7}$$

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 \frac{3}{7} for x_{2}.

$$-7x^{2}+3x=-7x\left(x-\frac{3}{7}\right)$$

Subtract \frac{3}{7} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-7x^{2}+3x=-7x\times \frac{-7x+3}{-7}$$

Cancel out 7, the greatest common factor in -7 and -7.

$$-7x^{2}+3x=x\left(-7x+3\right)$$

Evaluate

$$x\left(3-7x\right)$$