Solve 4x^2+3=x^2-7x

$$4 x ^ { 2 } + 3 = x ^ { 2 } - 7 x$$

Solve for x

$$x=\frac{\sqrt{13}-7}{6}\approx -0.565741454$$

$$x=\frac{-\sqrt{13}-7}{6}\approx -1.767591879$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$4x^{2}+3-x^{2}=-7x$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

Add 7x to both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\times 3}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$x=\frac{-7±\sqrt{49-36}}{2\times 3}$$

Add 49 to -36.

$$x=\frac{-7±\sqrt{13}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±\sqrt{13}}{6}$$

Now solve the equation x=\frac{-7±\sqrt{13}}{6} when ± is plus. Add -7 to \sqrt{13}.

$$x=\frac{\sqrt{13}-7}{6}$$

Now solve the equation x=\frac{-7±\sqrt{13}}{6} when ± is minus. Subtract \sqrt{13} from -7.

$$x=\frac{-\sqrt{13}-7}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{13}-7}{6} x=\frac{-\sqrt{13}-7}{6}$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$4x^{2}+3-x^{2}=-7x$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

Add 7x to both sides.

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

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

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

Divide -3 by 3.

$$x^{2}+\frac{7}{3}x=-1$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=-1+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=-1+\frac{49}{36}$$

Add -1 to \frac{49}{36}.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{13}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{13}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{13}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{\sqrt{13}}{6} x+\frac{7}{6}=-\frac{\sqrt{13}}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

$$x=\frac{\sqrt{13}-7}{6} x=\frac{-\sqrt{13}-7}{6}$$