Solve 3x^2+4x-5=3(-5)^2+4(-5)-5

$$3 x ^ { 2 } + 4 x - 5 = 3 ( - 5 ) ^ { 2 } + 4 ( - 5 ) - 5$$

Solve for x

$$x=-5$$

$$x = \frac{11}{3} = 3\frac{2}{3} \approx 3.666666667$$

Steps Using Factoring By Grouping

Calculate -5 to the power of 2 and get 25.

$$3x^{2}+4x-5=3\times 25+4\left(-5\right)-5$$

Multiply 3 and 25 to get 75.

$$3x^{2}+4x-5=75+4\left(-5\right)-5$$

Multiply 4 and -5 to get -20.

$$3x^{2}+4x-5=75-20-5$$

Subtract 20 from 75 to get 55.

$$3x^{2}+4x-5=55-5$$

Subtract 5 from 55 to get 50.

$$3x^{2}+4x-5=50$$

Subtract 50 from both sides.

$$3x^{2}+4x-5-50=0$$

Subtract 50 from -5 to get -55.

$$3x^{2}+4x-55=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-55. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\left(-55\right)=-165$$

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 -165.

$$-1,165 -3,55 -5,33 -11,15$$

Calculate the sum for each pair.

$$-1+165=164 -3+55=52 -5+33=28 -11+15=4$$

The solution is the pair that gives sum 4.

$$a=-11 b=15$$

Rewrite 3x^{2}+4x-55 as \left(3x^{2}-11x\right)+\left(15x-55\right).

$$\left(3x^{2}-11x\right)+\left(15x-55\right)$$

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

$$x\left(3x-11\right)+5\left(3x-11\right)$$

Factor out common term 3x-11 by using distributive property.

$$\left(3x-11\right)\left(x+5\right)$$

To find equation solutions, solve 3x-11=0 and x+5=0.

$$x=\frac{11}{3} x=-5$$

Steps Using the Quadratic Formula

Calculate -5 to the power of 2 and get 25.

$$3x^{2}+4x-5=3\times 25+4\left(-5\right)-5$$

Multiply 3 and 25 to get 75.

$$3x^{2}+4x-5=75+4\left(-5\right)-5$$

Multiply 4 and -5 to get -20.

$$3x^{2}+4x-5=75-20-5$$

Subtract 20 from 75 to get 55.

$$3x^{2}+4x-5=55-5$$

Subtract 5 from 55 to get 50.

$$3x^{2}+4x-5=50$$

Subtract 50 from both sides.

$$3x^{2}+4x-5-50=0$$

Subtract 50 from -5 to get -55.

$$3x^{2}+4x-55=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 4 for b, and -55 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 3\left(-55\right)}}{2\times 3}$$

Square 4.

$$x=\frac{-4±\sqrt{16-4\times 3\left(-55\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-4±\sqrt{16-12\left(-55\right)}}{2\times 3}$$

Multiply -12 times -55.

$$x=\frac{-4±\sqrt{16+660}}{2\times 3}$$

Add 16 to 660.

$$x=\frac{-4±\sqrt{676}}{2\times 3}$$

Take the square root of 676.

$$x=\frac{-4±26}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-4±26}{6}$$

Now solve the equation x=\frac{-4±26}{6} when ± is plus. Add -4 to 26.

$$x=\frac{22}{6}$$

Reduce the fraction \frac{22}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{11}{3}$$

Now solve the equation x=\frac{-4±26}{6} when ± is minus. Subtract 26 from -4.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

The equation is now solved.

$$x=\frac{11}{3} x=-5$$

Steps for Completing the Square

Calculate -5 to the power of 2 and get 25.

$$3x^{2}+4x-5=3\times 25+4\left(-5\right)-5$$

Multiply 3 and 25 to get 75.

$$3x^{2}+4x-5=75+4\left(-5\right)-5$$

Multiply 4 and -5 to get -20.

$$3x^{2}+4x-5=75-20-5$$

Subtract 20 from 75 to get 55.

$$3x^{2}+4x-5=55-5$$

Subtract 5 from 55 to get 50.

$$3x^{2}+4x-5=50$$

Add 5 to both sides.

$$3x^{2}+4x=50+5$$

Add 50 and 5 to get 55.

$$3x^{2}+4x=55$$

Divide both sides by 3.

$$\frac{3x^{2}+4x}{3}=\frac{55}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{4}{3}x=\frac{55}{3}$$

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

$$x^{2}+\frac{4}{3}x+\left(\frac{2}{3}\right)^{2}=\frac{55}{3}+\left(\frac{2}{3}\right)^{2}$$

Square \frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{55}{3}+\frac{4}{9}$$

Add \frac{55}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{4}{3}x+\frac{4}{9}=\frac{169}{9}$$

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

$$\left(x+\frac{2}{3}\right)^{2}=\frac{169}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{2}{3}\right)^{2}}=\sqrt{\frac{169}{9}}$$

Simplify.

$$x+\frac{2}{3}=\frac{13}{3} x+\frac{2}{3}=-\frac{13}{3}$$

Subtract \frac{2}{3} from both sides of the equation.

$$x=\frac{11}{3} x=-5$$