Solve for x

x=-1

x = \frac{3}{2} = 1\frac{1}{2} = 1.5

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a+b=-1 ab=2\left(-3\right)=-6

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

1,-6 2,-3

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 -6.

1-6=-5 2-3=-1

Calculate the sum for each pair.

a=-3 b=2

The solution is the pair that gives sum -1.

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

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

x\left(2x-3\right)+2x-3

Factor out x in 2x^{2}-3x.

\left(2x-3\right)\left(x+1\right)

Factor out common term 2x-3 by using distributive property.

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

To find equation solutions, solve 2x-3=0 and x+1=0.

2x^{2}-x-3=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{-\left(-1\right)±\sqrt{1-4\times 2\left(-3\right)}}{2\times 2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-1\right)±\sqrt{1-8\left(-3\right)}}{2\times 2}

Multiply -4 times 2.

x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\times 2}

Multiply -8 times -3.

x=\frac{-\left(-1\right)±\sqrt{25}}{2\times 2}

Add 1 to 24.

x=\frac{-\left(-1\right)±5}{2\times 2}

Take the square root of 25.

x=\frac{1±5}{2\times 2}

The opposite of -1 is 1.

x=\frac{1±5}{4}

Multiply 2 times 2.

x=\frac{6}{4}

Now solve the equation x=\frac{1±5}{4} when ± is plus. Add 1 to 5.

x=\frac{3}{2}

Reduce the fraction \frac{6}{4} to lowest terms by extracting and canceling out 2.

x=\frac{-4}{4}

Now solve the equation x=\frac{1±5}{4} when ± is minus. Subtract 5 from 1.

x=-1

Divide -4 by 4.

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

The equation is now solved.

2x^{2}-x-3=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.

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

Add 3 to both sides of the equation.

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

Subtracting -3 from itself leaves 0.

2x^{2}-x=3

Subtract -3 from 0.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide -1 by 2.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{3}{2}+\frac{1}{16}

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}

Add \frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

\left(x-\frac{1}{4}\right)^{2}=\frac{25}{16}

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. 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{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{1}{4} to both sides of the equation.

x ^ 2 -\frac{1}{2}x -\frac{3}{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 2

r + s = \frac{1}{2} rs = -\frac{3}{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{1}{4} - u s = \frac{1}{4} + u

Two numbers r and s sum up to \frac{1}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{2} = \frac{1}{4}. 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{1}{4} - u) (\frac{1}{4} + u) = -\frac{3}{2}

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}

\frac{1}{16} - u^2 = -\frac{3}{2}

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

-u^2 = -\frac{3}{2}-\frac{1}{16} = -\frac{25}{16}

Simplify the expression by subtracting \frac{1}{16} on both sides

u^2 = \frac{25}{16} u = \pm\sqrt{\frac{25}{16}} = \pm \frac{5}{4}

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

r =\frac{1}{4} - \frac{5}{4} = -1 s = \frac{1}{4} + \frac{5}{4} = 1.500

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

Examples

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

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

Linear equation

y = 3x + 4

Arithmetic

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]

Simultaneous equation

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

Differentiation

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

Integration

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

Limits

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