Solve 49x^2-140x+120=0

$$49 { x }^{ 2 } -140x+120 = 0$$

Solve for x (complex solution)

$$x=\frac{10+2\sqrt{5}i}{7}\approx 1.428571429+0.638876565i$$

$$x=\frac{-2\sqrt{5}i+10}{7}\approx 1.428571429-0.638876565i$$

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.

$$49x^{2}-140x+120=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, -140 for b, and 120 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-140\right)±\sqrt{\left(-140\right)^{2}-4\times 49\times 120}}{2\times 49}$$

Square -140.

$$x=\frac{-\left(-140\right)±\sqrt{19600-4\times 49\times 120}}{2\times 49}$$

Multiply -4 times 49.

$$x=\frac{-\left(-140\right)±\sqrt{19600-196\times 120}}{2\times 49}$$

Multiply -196 times 120.

$$x=\frac{-\left(-140\right)±\sqrt{19600-23520}}{2\times 49}$$

Add 19600 to -23520.

$$x=\frac{-\left(-140\right)±\sqrt{-3920}}{2\times 49}$$

Take the square root of -3920.

$$x=\frac{-\left(-140\right)±28\sqrt{5}i}{2\times 49}$$

The opposite of -140 is 140.

$$x=\frac{140±28\sqrt{5}i}{2\times 49}$$

Multiply 2 times 49.

$$x=\frac{140±28\sqrt{5}i}{98}$$

Now solve the equation x=\frac{140±28\sqrt{5}i}{98} when ± is plus. Add 140 to 28i\sqrt{5}.

$$x=\frac{140+28\sqrt{5}i}{98}$$

Divide 140+28i\sqrt{5} by 98.

$$x=\frac{10+2\sqrt{5}i}{7}$$

Now solve the equation x=\frac{140±28\sqrt{5}i}{98} when ± is minus. Subtract 28i\sqrt{5} from 140.

$$x=\frac{-28\sqrt{5}i+140}{98}$$

Divide 140-28i\sqrt{5} by 98.

$$x=\frac{-2\sqrt{5}i+10}{7}$$

The equation is now solved.

$$x=\frac{10+2\sqrt{5}i}{7} x=\frac{-2\sqrt{5}i+10}{7}$$

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.

$$49x^{2}-140x+120=0$$

Subtract 120 from both sides of the equation.

$$49x^{2}-140x+120-120=-120$$

Subtracting 120 from itself leaves 0.

$$49x^{2}-140x=-120$$

Divide both sides by 49.

$$\frac{49x^{2}-140x}{49}=-\frac{120}{49}$$

Dividing by 49 undoes the multiplication by 49.

$$x^{2}+\left(-\frac{140}{49}\right)x=-\frac{120}{49}$$

Reduce the fraction \frac{-140}{49} to lowest terms by extracting and canceling out 7.

$$x^{2}-\frac{20}{7}x=-\frac{120}{49}$$

Divide -\frac{20}{7}, the coefficient of the x term, by 2 to get -\frac{10}{7}. Then add the square of -\frac{10}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{20}{7}x+\left(-\frac{10}{7}\right)^{2}=-\frac{120}{49}+\left(-\frac{10}{7}\right)^{2}$$

Square -\frac{10}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{20}{7}x+\frac{100}{49}=\frac{-120+100}{49}$$

Add -\frac{120}{49} to \frac{100}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{20}{7}x+\frac{100}{49}=-\frac{20}{49}$$

Factor x^{2}-\frac{20}{7}x+\frac{100}{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}.

$$\left(x-\frac{10}{7}\right)^{2}=-\frac{20}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{10}{7}\right)^{2}}=\sqrt{-\frac{20}{49}}$$

Simplify.

$$x-\frac{10}{7}=\frac{2\sqrt{5}i}{7} x-\frac{10}{7}=-\frac{2\sqrt{5}i}{7}$$

Add \frac{10}{7} to both sides of the equation.

$$x=\frac{10+2\sqrt{5}i}{7} x=\frac{-2\sqrt{5}i+10}{7}$$