Solve 3v=10v^2-4

$$3 v = 10 v ^ { 2 } - 4$$

Solve for v

$$v=\frac{4}{5}=0.8$$

$$v=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

Subtract 10v^{2} from both sides.

$$3v-10v^{2}=-4$$

Add 4 to both sides.

$$3v-10v^{2}+4=0$$

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

$$-10v^{2}+3v+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -10v^{2}+av+bv+4. To find a and b, set up a system to be solved.

$$a+b=3 ab=-10\times 4=-40$$

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 3.

$$a=8 b=-5$$

Rewrite -10v^{2}+3v+4 as \left(-10v^{2}+8v\right)+\left(-5v+4\right).

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

Factor out 2v in -10v^{2}+8v.

$$2v\left(-5v+4\right)-5v+4$$

Factor out common term -5v+4 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract 10v^{2} from both sides.

$$3v-10v^{2}=-4$$

Add 4 to both sides.

$$3v-10v^{2}+4=0$$

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.

$$-10v^{2}+3v+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 3 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-3±\sqrt{3^{2}-4\left(-10\right)\times 4}}{2\left(-10\right)}$$

Square 3.

$$v=\frac{-3±\sqrt{9-4\left(-10\right)\times 4}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$v=\frac{-3±\sqrt{9+40\times 4}}{2\left(-10\right)}$$

Multiply 40 times 4.

$$v=\frac{-3±\sqrt{9+160}}{2\left(-10\right)}$$

Add 9 to 160.

$$v=\frac{-3±\sqrt{169}}{2\left(-10\right)}$$

Take the square root of 169.

$$v=\frac{-3±13}{2\left(-10\right)}$$

Multiply 2 times -10.

$$v=\frac{-3±13}{-20}$$

Now solve the equation v=\frac{-3±13}{-20} when ± is plus. Add -3 to 13.

$$v=\frac{10}{-20}$$

Reduce the fraction \frac{10}{-20} to lowest terms by extracting and canceling out 10.

$$v=-\frac{1}{2}$$

Now solve the equation v=\frac{-3±13}{-20} when ± is minus. Subtract 13 from -3.

$$v=-\frac{16}{-20}$$

Reduce the fraction \frac{-16}{-20} to lowest terms by extracting and canceling out 4.

$$v=\frac{4}{5}$$

The equation is now solved.

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

Steps for Completing the Square

Subtract 10v^{2} from both sides.

$$3v-10v^{2}=-4$$

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.

$$-10v^{2}+3v=-4$$

Divide both sides by -10.

$$\frac{-10v^{2}+3v}{-10}=-\frac{4}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$v^{2}+\frac{3}{-10}v=-\frac{4}{-10}$$

Divide 3 by -10.

$$v^{2}-\frac{3}{10}v=-\frac{4}{-10}$$

Reduce the fraction \frac{-4}{-10} to lowest terms by extracting and canceling out 2.

$$v^{2}-\frac{3}{10}v=\frac{2}{5}$$

Divide -\frac{3}{10}, the coefficient of the x term, by 2 to get -\frac{3}{20}. Then add the square of -\frac{3}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}-\frac{3}{10}v+\left(-\frac{3}{20}\right)^{2}=\frac{2}{5}+\left(-\frac{3}{20}\right)^{2}$$

Square -\frac{3}{20} by squaring both the numerator and the denominator of the fraction.

$$v^{2}-\frac{3}{10}v+\frac{9}{400}=\frac{2}{5}+\frac{9}{400}$$

Add \frac{2}{5} to \frac{9}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$v^{2}-\frac{3}{10}v+\frac{9}{400}=\frac{169}{400}$$

Factor v^{2}-\frac{3}{10}v+\frac{9}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(v-\frac{3}{20}\right)^{2}=\frac{169}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-\frac{3}{20}\right)^{2}}=\sqrt{\frac{169}{400}}$$

Simplify.

$$v-\frac{3}{20}=\frac{13}{20} v-\frac{3}{20}=-\frac{13}{20}$$

Add \frac{3}{20} to both sides of the equation.

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