Solve for x

x=-2<br/>x=-1

$x=−2$

$x=−1$

$x=−1$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Direct Factoring Method

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x^{2}+3x+2=0

Divide both sides by 3.

a+b=3 ab=1\times 2=2

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

a=1 b=2

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

\left(x^{2}+x\right)+\left(2x+2\right)

Rewrite x^{2}+3x+2 as \left(x^{2}+x\right)+\left(2x+2\right).

x\left(x+1\right)+2\left(x+1\right)

Factor out x in the first and 2 in the second group.

\left(x+1\right)\left(x+2\right)

Factor out common term x+1 by using distributive property.

x=-1 x=-2

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

3x^{2}+9x+6=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.

x=\frac{-9±\sqrt{9^{2}-4\times 3\times 6}}{2\times 3}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 9 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-9±\sqrt{81-4\times 3\times 6}}{2\times 3}

Square 9.

x=\frac{-9±\sqrt{81-12\times 6}}{2\times 3}

Multiply -4 times 3.

x=\frac{-9±\sqrt{81-72}}{2\times 3}

Multiply -12 times 6.

x=\frac{-9±\sqrt{9}}{2\times 3}

Add 81 to -72.

x=\frac{-9±3}{2\times 3}

Take the square root of 9.

x=\frac{-9±3}{6}

Multiply 2 times 3.

x=\frac{-6}{6}

Now solve the equation x=\frac{-9±3}{6} when ± is plus. Add -9 to 3.

x=-1

Divide -6 by 6.

x=\frac{-12}{6}

Now solve the equation x=\frac{-9±3}{6} when ± is minus. Subtract 3 from -9.

x=-2

Divide -12 by 6.

x=-1 x=-2

The equation is now solved.

3x^{2}+9x+6=0

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.

3x^{2}+9x+6-6=-6

Subtract 6 from both sides of the equation.

3x^{2}+9x=-6

Subtracting 6 from itself leaves 0.

\frac{3x^{2}+9x}{3}=\frac{-6}{3}

Divide both sides by 3.

x^{2}+\frac{9}{3}x=\frac{-6}{3}

Dividing by 3 undoes the multiplication by 3.

x^{2}+3x=\frac{-6}{3}

Divide 9 by 3.

x^{2}+3x=-2

Divide -6 by 3.

x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-2+\left(\frac{3}{2}\right)^{2}

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}=1.5. Then add the square of \frac{3}{2}=1.5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{2}=1.5 by squaring both the numerator and the denominator of the fraction.

x^{2}+3x+\frac{9}{4}=\frac{1}{4}

Add -2 to \frac{9}{4}=2.25.

\left(x+\frac{3}{2}\right)^{2}=\frac{1}{4}

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

\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}

Take the square root of both sides of the equation.

x+\frac{3}{2}=\frac{1}{2} x+\frac{3}{2}=-\frac{1}{2}

Simplify.

x=-1 x=-2

Subtract \frac{3}{2}=1.5 from both sides of the equation.

x ^ 2 +3x +2 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 3

r + s = -3 rs = 2

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = -\frac{3}{2} - u s = -\frac{3}{2} + u

Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(-\frac{3}{2} - u) (-\frac{3}{2} + u) = 2

To solve for unknown quantity u, substitute these in the product equation rs = 2

\frac{9}{4} - u^2 = 2

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = 2-\frac{9}{4} = -\frac{1}{4}

Simplify the expression by subtracting \frac{9}{4} on both sides

u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r =-\frac{3}{2} - \frac{1}{2} = -2 s = -\frac{3}{2} + \frac{1}{2} = -1

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

Examples

Quadratic equation

{ x } ^ { 2 } - 4 x - 5 = 0

$x_{2}−4x−5=0$

Trigonometry

4 \sin \theta \cos \theta = 2 \sin \theta

$4sinθcosθ=2sinθ$

Linear equation

y = 3x + 4

$y=3x+4$

Arithmetic

699 * 533

$699∗533$

Matrix

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { - 1 } & { 1 } & { 5 } \end{array} \right]

$[25 34 ][2−1 01 35 ]$

Simultaneous equation

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

${8x+2y=467x+3y=47 $

Differentiation

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

$dxd (x−5)(3x_{2}−2) $

Integration

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

$∫_{0}xe_{−x_{2}}dx$

Limits

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

$x→−3lim x_{2}+2x−3x_{2}−9 $