Solve 2a^2-20a+100=178

$$2 a ^ { 2 } - 20 a + 100 = 178$$

Solve for a

$$a=-3$$

$$a=13$$

Steps Using Factoring By Grouping

Subtract 178 from both sides.

$$2a^{2}-20a+100-178=0$$

Subtract 178 from 100 to get -78.

$$2a^{2}-20a-78=0$$

Divide both sides by 2.

$$a^{2}-10a-39=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as a^{2}+aa+ba-39. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-39\right)=-39$$

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 -39.

$$1,-39 3,-13$$

Calculate the sum for each pair.

$$1-39=-38 3-13=-10$$

The solution is the pair that gives sum -10.

$$a=-13 b=3$$

Rewrite a^{2}-10a-39 as \left(a^{2}-13a\right)+\left(3a-39\right).

$$\left(a^{2}-13a\right)+\left(3a-39\right)$$

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

$$a\left(a-13\right)+3\left(a-13\right)$$

Factor out common term a-13 by using distributive property.

$$\left(a-13\right)\left(a+3\right)$$

To find equation solutions, solve a-13=0 and a+3=0.

$$a=13 a=-3$$

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.

$$2a^{2}-20a+100=178$$

Subtract 178 from both sides of the equation.

$$2a^{2}-20a+100-178=178-178$$

Subtracting 178 from itself leaves 0.

$$2a^{2}-20a+100-178=0$$

Subtract 178 from 100.

$$2a^{2}-20a-78=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -20 for b, and -78 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 2\left(-78\right)}}{2\times 2}$$

Square -20.

$$a=\frac{-\left(-20\right)±\sqrt{400-4\times 2\left(-78\right)}}{2\times 2}$$

Multiply -4 times 2.

$$a=\frac{-\left(-20\right)±\sqrt{400-8\left(-78\right)}}{2\times 2}$$

Multiply -8 times -78.

$$a=\frac{-\left(-20\right)±\sqrt{400+624}}{2\times 2}$$

Add 400 to 624.

$$a=\frac{-\left(-20\right)±\sqrt{1024}}{2\times 2}$$

Take the square root of 1024.

$$a=\frac{-\left(-20\right)±32}{2\times 2}$$

The opposite of -20 is 20.

$$a=\frac{20±32}{2\times 2}$$

Multiply 2 times 2.

$$a=\frac{20±32}{4}$$

Now solve the equation a=\frac{20±32}{4} when ± is plus. Add 20 to 32.

$$a=\frac{52}{4}$$

Divide 52 by 4.

$$a=13$$

Now solve the equation a=\frac{20±32}{4} when ± is minus. Subtract 32 from 20.

$$a=-\frac{12}{4}$$

Divide -12 by 4.

$$a=-3$$

The equation is now solved.

$$a=13 a=-3$$

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.

$$2a^{2}-20a+100=178$$

Subtract 100 from both sides of the equation.

$$2a^{2}-20a+100-100=178-100$$

Subtracting 100 from itself leaves 0.

$$2a^{2}-20a=178-100$$

Subtract 100 from 178.

$$2a^{2}-20a=78$$

Divide both sides by 2.

$$\frac{2a^{2}-20a}{2}=\frac{78}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$a^{2}+\left(-\frac{20}{2}\right)a=\frac{78}{2}$$

Divide -20 by 2.

$$a^{2}-10a=\frac{78}{2}$$

Divide 78 by 2.

$$a^{2}-10a=39$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$a^{2}-10a+\left(-5\right)^{2}=39+\left(-5\right)^{2}$$

Square -5.

$$a^{2}-10a+25=39+25$$

Add 39 to 25.

$$a^{2}-10a+25=64$$

Factor a^{2}-10a+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(a-5\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a-5\right)^{2}}=\sqrt{64}$$

Simplify.

$$a-5=8 a-5=-8$$

Add 5 to both sides of the equation.

$$a=13 a=-3$$