Solve 2s^2-70=4s

$$2 s ^ { 2 } - 70 = 4 s$$

Solve for s

$$s=-5$$

$$s=7$$

Steps Using Factoring By Grouping

Subtract 4s from both sides.

$$2s^{2}-70-4s=0$$

Divide both sides by 2.

$$s^{2}-35-2s=0$$

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

$$s^{2}-2s-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as s^{2}+as+bs-35. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-35\right)=-35$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -35.

$$1,-35 5,-7$$

Calculate the sum for each pair.

$$1-35=-34 5-7=-2$$

The solution is the pair that gives sum -2.

$$a=-7 b=5$$

Rewrite s^{2}-2s-35 as \left(s^{2}-7s\right)+\left(5s-35\right).

$$\left(s^{2}-7s\right)+\left(5s-35\right)$$

Factor out s in the first and 5 in the second group.

$$s\left(s-7\right)+5\left(s-7\right)$$

Factor out common term s-7 by using distributive property.

$$\left(s-7\right)\left(s+5\right)$$

To find equation solutions, solve s-7=0 and s+5=0.

$$s=7 s=-5$$

Steps Using the Quadratic Formula

Subtract 4s from both sides.

$$2s^{2}-70-4s=0$$

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.

$$2s^{2}-4s-70=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and -70 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$s=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\left(-70\right)}}{2\times 2}$$

Square -4.

$$s=\frac{-\left(-4\right)±\sqrt{16-4\times 2\left(-70\right)}}{2\times 2}$$

Multiply -4 times 2.

$$s=\frac{-\left(-4\right)±\sqrt{16-8\left(-70\right)}}{2\times 2}$$

Multiply -8 times -70.

$$s=\frac{-\left(-4\right)±\sqrt{16+560}}{2\times 2}$$

Add 16 to 560.

$$s=\frac{-\left(-4\right)±\sqrt{576}}{2\times 2}$$

Take the square root of 576.

$$s=\frac{-\left(-4\right)±24}{2\times 2}$$

The opposite of -4 is 4.

$$s=\frac{4±24}{2\times 2}$$

Multiply 2 times 2.

$$s=\frac{4±24}{4}$$

Now solve the equation s=\frac{4±24}{4} when ± is plus. Add 4 to 24.

$$s=\frac{28}{4}$$

Divide 28 by 4.

$$s=7$$

Now solve the equation s=\frac{4±24}{4} when ± is minus. Subtract 24 from 4.

$$s=-\frac{20}{4}$$

Divide -20 by 4.

$$s=-5$$

The equation is now solved.

$$s=7 s=-5$$

Steps for Completing the Square

Subtract 4s from both sides.

$$2s^{2}-70-4s=0$$

Add 70 to both sides. Anything plus zero gives itself.

$$2s^{2}-4s=70$$

Divide both sides by 2.

$$\frac{2s^{2}-4s}{2}=\frac{70}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$s^{2}+\left(-\frac{4}{2}\right)s=\frac{70}{2}$$

Divide -4 by 2.

$$s^{2}-2s=\frac{70}{2}$$

Divide 70 by 2.

$$s^{2}-2s=35$$

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.

$$s^{2}-2s+1=35+1$$

Add 35 to 1.

$$s^{2}-2s+1=36$$

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}.

$$\left(s-1\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(s-1\right)^{2}}=\sqrt{36}$$

Simplify.

$$s-1=6 s-1=-6$$

Add 1 to both sides of the equation.

$$s=7 s=-5$$