Solve 2x^2*120+6x*30+45=195

$$2 x ^ { 2 } \cdot 120 + 6 x \cdot 30 + 45 = 195$$

Solve for x

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Multiply 2 and 120 to get 240.

$$240x^{2}+6x\times 30+45=195$$

Multiply 6 and 30 to get 180.

$$240x^{2}+180x+45=195$$

Subtract 195 from both sides.

$$240x^{2}+180x+45-195=0$$

Subtract 195 from 45 to get -150.

$$240x^{2}+180x-150=0$$

Divide both sides by 30.

$$8x^{2}+6x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=6 ab=8\left(-5\right)=-40$$

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. List all such integer pairs that give product -40.

$$-1,40 -2,20 -4,10 -5,8$$

Calculate the sum for each pair.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

The solution is the pair that gives sum 6.

$$a=-4 b=10$$

Rewrite 8x^{2}+6x-5 as \left(8x^{2}-4x\right)+\left(10x-5\right).

$$\left(8x^{2}-4x\right)+\left(10x-5\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(2x-1\right)+5\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(4x+5\right)$$

To find equation solutions, solve 2x-1=0 and 4x+5=0.

$$x=\frac{1}{2} x=-\frac{5}{4}$$

Steps Using the Quadratic Formula

Multiply 2 and 120 to get 240.

$$240x^{2}+6x\times 30+45=195$$

Multiply 6 and 30 to get 180.

$$240x^{2}+180x+45=195$$

Subtract 195 from both sides.

$$240x^{2}+180x+45-195=0$$

Subtract 195 from 45 to get -150.

$$240x^{2}+180x-150=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 240 for a, 180 for b, and -150 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-180±\sqrt{180^{2}-4\times 240\left(-150\right)}}{2\times 240}$$

Square 180.

$$x=\frac{-180±\sqrt{32400-4\times 240\left(-150\right)}}{2\times 240}$$

Multiply -4 times 240.

$$x=\frac{-180±\sqrt{32400-960\left(-150\right)}}{2\times 240}$$

Multiply -960 times -150.

$$x=\frac{-180±\sqrt{32400+144000}}{2\times 240}$$

Add 32400 to 144000.

$$x=\frac{-180±\sqrt{176400}}{2\times 240}$$

Take the square root of 176400.

$$x=\frac{-180±420}{2\times 240}$$

Multiply 2 times 240.

$$x=\frac{-180±420}{480}$$

Now solve the equation x=\frac{-180±420}{480} when ± is plus. Add -180 to 420.

$$x=\frac{240}{480}$$

Reduce the fraction \frac{240}{480} to lowest terms by extracting and canceling out 240.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-180±420}{480} when ± is minus. Subtract 420 from -180.

$$x=-\frac{600}{480}$$

Reduce the fraction \frac{-600}{480} to lowest terms by extracting and canceling out 120.

$$x=-\frac{5}{4}$$

The equation is now solved.

$$x=\frac{1}{2} x=-\frac{5}{4}$$

Steps for Completing the Square

Multiply 2 and 120 to get 240.

$$240x^{2}+6x\times 30+45=195$$

Multiply 6 and 30 to get 180.

$$240x^{2}+180x+45=195$$

Subtract 45 from both sides.

$$240x^{2}+180x=195-45$$

Subtract 45 from 195 to get 150.

$$240x^{2}+180x=150$$

Divide both sides by 240.

$$\frac{240x^{2}+180x}{240}=\frac{150}{240}$$

Dividing by 240 undoes the multiplication by 240.

$$x^{2}+\frac{180}{240}x=\frac{150}{240}$$

Reduce the fraction \frac{180}{240} to lowest terms by extracting and canceling out 60.

$$x^{2}+\frac{3}{4}x=\frac{150}{240}$$

Reduce the fraction \frac{150}{240} to lowest terms by extracting and canceling out 30.

$$x^{2}+\frac{3}{4}x=\frac{5}{8}$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=\frac{5}{8}+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{5}{8}+\frac{9}{64}$$

Add \frac{5}{8} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{49}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{7}{8} x+\frac{3}{8}=-\frac{7}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=\frac{1}{2} x=-\frac{5}{4}$$