Solve s^2-3=-2s

$$s ^ { 2 } - 3 = - 2 s$$

Solve for s

$$s=-3$$

$$s=1$$

Steps Using Factoring

Add 2s to both sides.

$$s^{2}-3+2s=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$s^{2}+2s-3=0$$

To solve the equation, factor s^{2}+2s-3 using formula s^{2}+\left(a+b\right)s+ab=\left(s+a\right)\left(s+b\right). To find a and b, set up a system to be solved.

$$a+b=2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

Rewrite factored expression \left(s+a\right)\left(s+b\right) using the obtained values.

$$\left(s-1\right)\left(s+3\right)$$

To find equation solutions, solve s-1=0 and s+3=0.

$$s=1 s=-3$$

Steps Using Factoring By Grouping

Add 2s to both sides.

$$s^{2}-3+2s=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$s^{2}+2s-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as s^{2}+as+bs-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

Rewrite s^{2}+2s-3 as \left(s^{2}-s\right)+\left(3s-3\right).

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

Factor out s in the first and 3 in the second group.

$$s\left(s-1\right)+3\left(s-1\right)$$

Factor out common term s-1 by using distributive property.

$$\left(s-1\right)\left(s+3\right)$$

To find equation solutions, solve s-1=0 and s+3=0.

$$s=1 s=-3$$

Steps Using the Quadratic Formula

Add 2s to both sides.

$$s^{2}-3+2s=0$$

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.

$$s^{2}+2s-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$s=\frac{-2±\sqrt{2^{2}-4\left(-3\right)}}{2}$$

Square 2.

$$s=\frac{-2±\sqrt{4-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$s=\frac{-2±\sqrt{4+12}}{2}$$

Add 4 to 12.

$$s=\frac{-2±\sqrt{16}}{2}$$

Take the square root of 16.

$$s=\frac{-2±4}{2}$$

Now solve the equation s=\frac{-2±4}{2} when ± is plus. Add -2 to 4.

$$s=\frac{2}{2}$$

Divide 2 by 2.

$$s=1$$

Now solve the equation s=\frac{-2±4}{2} when ± is minus. Subtract 4 from -2.

$$s=-\frac{6}{2}$$

Divide -6 by 2.

$$s=-3$$

The equation is now solved.

$$s=1 s=-3$$

Steps for Completing the Square

Add 2s to both sides.

$$s^{2}-3+2s=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$s^{2}+2s=3$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$s^{2}+2s+1^{2}=3+1^{2}$$

Square 1.

$$s^{2}+2s+1=3+1$$

Add 3 to 1.

$$s^{2}+2s+1=4$$

Factor s^{2}+2s+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(s+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(s+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$s+1=2 s+1=-2$$

Subtract 1 from both sides of the equation.

$$s=1 s=-3$$