Solve for x
x=-\frac{1}{3}\approx -0.333333333
x = \frac{6}{5} = 1\frac{1}{5} = 1.2
Graph
Share
Copied to clipboard
15x^{2}-13x-6=0
Divide both sides by 3.
a+b=-13 ab=15\left(-6\right)=-90
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 15x^{2}+ax+bx-6. To find a and b, set up a system to be solved.
1,-90 2,-45 3,-30 5,-18 6,-15 9,-10
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 -90.
1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1
Calculate the sum for each pair.
a=-18 b=5
The solution is the pair that gives sum -13.
\left(15x^{2}-18x\right)+\left(5x-6\right)
Rewrite 15x^{2}-13x-6 as \left(15x^{2}-18x\right)+\left(5x-6\right).
3x\left(5x-6\right)+5x-6
Factor out 3x in 15x^{2}-18x.
\left(5x-6\right)\left(3x+1\right)
Factor out common term 5x-6 by using distributive property.
x=\frac{6}{5} x=-\frac{1}{3}
To find equation solutions, solve 5x-6=0 and 3x+1=0.
45x^{2}-39x-18=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(-39\right)±\sqrt{\left(-39\right)^{2}-4\times 45\left(-18\right)}}{2\times 45}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 45 for a, -39 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-39\right)±\sqrt{1521-4\times 45\left(-18\right)}}{2\times 45}
Square -39.
x=\frac{-\left(-39\right)±\sqrt{1521-180\left(-18\right)}}{2\times 45}
Multiply -4 times 45.
x=\frac{-\left(-39\right)±\sqrt{1521+3240}}{2\times 45}
Multiply -180 times -18.
x=\frac{-\left(-39\right)±\sqrt{4761}}{2\times 45}
Add 1521 to 3240.
x=\frac{-\left(-39\right)±69}{2\times 45}
Take the square root of 4761.
x=\frac{39±69}{2\times 45}
The opposite of -39 is 39.
x=\frac{39±69}{90}
Multiply 2 times 45.
x=\frac{108}{90}
Now solve the equation x=\frac{39±69}{90} when ± is plus. Add 39 to 69.
x=\frac{6}{5}
Reduce the fraction \frac{108}{90} to lowest terms by extracting and canceling out 18.
x=-\frac{30}{90}
Now solve the equation x=\frac{39±69}{90} when ± is minus. Subtract 69 from 39.
x=-\frac{1}{3}
Reduce the fraction \frac{-30}{90} to lowest terms by extracting and canceling out 30.
x=\frac{6}{5} x=-\frac{1}{3}
The equation is now solved.
45x^{2}-39x-18=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.
45x^{2}-39x-18-\left(-18\right)=-\left(-18\right)
Add 18 to both sides of the equation.
45x^{2}-39x=-\left(-18\right)
Subtracting -18 from itself leaves 0.
45x^{2}-39x=18
Subtract -18 from 0.
\frac{45x^{2}-39x}{45}=\frac{18}{45}
Divide both sides by 45.
x^{2}+\left(-\frac{39}{45}\right)x=\frac{18}{45}
Dividing by 45 undoes the multiplication by 45.
x^{2}-\frac{13}{15}x=\frac{18}{45}
Reduce the fraction \frac{-39}{45} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{13}{15}x=\frac{2}{5}
Reduce the fraction \frac{18}{45} to lowest terms by extracting and canceling out 9.
x^{2}-\frac{13}{15}x+\left(-\frac{13}{30}\right)^{2}=\frac{2}{5}+\left(-\frac{13}{30}\right)^{2}
Divide -\frac{13}{15}, the coefficient of the x term, by 2 to get -\frac{13}{30}. Then add the square of -\frac{13}{30} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{13}{15}x+\frac{169}{900}=\frac{2}{5}+\frac{169}{900}
Square -\frac{13}{30} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{13}{15}x+\frac{169}{900}=\frac{529}{900}
Add \frac{2}{5} to \frac{169}{900} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{13}{30}\right)^{2}=\frac{529}{900}
Factor x^{2}-\frac{13}{15}x+\frac{169}{900}. 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{13}{30}\right)^{2}}=\sqrt{\frac{529}{900}}
Take the square root of both sides of the equation.
x-\frac{13}{30}=\frac{23}{30} x-\frac{13}{30}=-\frac{23}{30}
Simplify.
x=\frac{6}{5} x=-\frac{1}{3}
Add \frac{13}{30} to both sides of the equation.
x ^ 2 -\frac{13}{15}x -\frac{2}{5} = 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 45
r + s = \frac{13}{15} rs = -\frac{2}{5}
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{13}{30} - u s = \frac{13}{30} + u
Two numbers r and s sum up to \frac{13}{15} exactly when the average of the two numbers is \frac{1}{2}*\frac{13}{15} = \frac{13}{30}. 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{13}{30} - u) (\frac{13}{30} + u) = -\frac{2}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{5}
\frac{169}{900} - u^2 = -\frac{2}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{5}-\frac{169}{900} = -\frac{529}{900}
Simplify the expression by subtracting \frac{169}{900} on both sides
u^2 = \frac{529}{900} u = \pm\sqrt{\frac{529}{900}} = \pm \frac{23}{30}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{13}{30} - \frac{23}{30} = -0.333 s = \frac{13}{30} + \frac{23}{30} = 1.200
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
Trigonometry
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}