Solve 49x^2-28x+4=12

$$49 { x }^{ 2 } -28x+4=12$$

Solve for x

$$x=\frac{2\sqrt{3}+2}{7}\approx 0.780585945$$

$$x=\frac{2-2\sqrt{3}}{7}\approx -0.209157374$$

Steps Using the Quadratic Formula

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.

$$49x^{2}-28x+4=12$$

Subtract 12 from both sides of the equation.

$$49x^{2}-28x+4-12=12-12$$

Subtracting 12 from itself leaves 0.

$$49x^{2}-28x+4-12=0$$

Subtract 12 from 4.

$$49x^{2}-28x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, -28 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 49\left(-8\right)}}{2\times 49}$$

Square -28.

$$x=\frac{-\left(-28\right)±\sqrt{784-4\times 49\left(-8\right)}}{2\times 49}$$

Multiply -4 times 49.

$$x=\frac{-\left(-28\right)±\sqrt{784-196\left(-8\right)}}{2\times 49}$$

Multiply -196 times -8.

$$x=\frac{-\left(-28\right)±\sqrt{784+1568}}{2\times 49}$$

Add 784 to 1568.

$$x=\frac{-\left(-28\right)±\sqrt{2352}}{2\times 49}$$

Take the square root of 2352.

$$x=\frac{-\left(-28\right)±28\sqrt{3}}{2\times 49}$$

The opposite of -28 is 28.

$$x=\frac{28±28\sqrt{3}}{2\times 49}$$

Multiply 2 times 49.

$$x=\frac{28±28\sqrt{3}}{98}$$

Now solve the equation x=\frac{28±28\sqrt{3}}{98} when ± is plus. Add 28 to 28\sqrt{3}.

$$x=\frac{28\sqrt{3}+28}{98}$$

Divide 28+28\sqrt{3} by 98.

$$x=\frac{2\sqrt{3}+2}{7}$$

Now solve the equation x=\frac{28±28\sqrt{3}}{98} when ± is minus. Subtract 28\sqrt{3} from 28.

$$x=\frac{28-28\sqrt{3}}{98}$$

Divide 28-28\sqrt{3} by 98.

$$x=\frac{2-2\sqrt{3}}{7}$$

The equation is now solved.

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

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$49x^{2}-28x+4=12$$

Subtract 4 from both sides of the equation.

$$49x^{2}-28x+4-4=12-4$$

Subtracting 4 from itself leaves 0.

$$49x^{2}-28x=12-4$$

Subtract 4 from 12.

$$49x^{2}-28x=8$$

Divide both sides by 49.

$$\frac{49x^{2}-28x}{49}=\frac{8}{49}$$

Dividing by 49 undoes the multiplication by 49.

$$x^{2}+\left(-\frac{28}{49}\right)x=\frac{8}{49}$$

Reduce the fraction \frac{-28}{49} to lowest terms by extracting and canceling out 7.

$$x^{2}-\frac{4}{7}x=\frac{8}{49}$$

Divide -\frac{4}{7}, the coefficient of the x term, by 2 to get -\frac{2}{7}. Then add the square of -\frac{2}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{7}x+\left(-\frac{2}{7}\right)^{2}=\frac{8}{49}+\left(-\frac{2}{7}\right)^{2}$$

Square -\frac{2}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{7}x+\frac{4}{49}=\frac{8+4}{49}$$

Add \frac{8}{49} to \frac{4}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{7}x+\frac{4}{49}=\frac{12}{49}$$

Factor x^{2}-\frac{4}{7}x+\frac{4}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{7}\right)^{2}=\frac{12}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{7}\right)^{2}}=\sqrt{\frac{12}{49}}$$

Simplify.

$$x-\frac{2}{7}=\frac{2\sqrt{3}}{7} x-\frac{2}{7}=-\frac{2\sqrt{3}}{7}$$

Add \frac{2}{7} to both sides of the equation.

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