Solve for x
x = \frac{\sqrt{43} + 5}{3} \approx 3.852479508
x=\frac{5-\sqrt{43}}{3}\approx -0.519146175
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3x^{2}-10x-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{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\left(-6\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\left(-6\right)}}{2\times 3}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-12\left(-6\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-10\right)±\sqrt{100+72}}{2\times 3}
Multiply -12 times -6.
x=\frac{-\left(-10\right)±\sqrt{172}}{2\times 3}
Add 100 to 72.
x=\frac{-\left(-10\right)±2\sqrt{43}}{2\times 3}
Take the square root of 172.
x=\frac{10±2\sqrt{43}}{2\times 3}
The opposite of -10 is 10.
x=\frac{10±2\sqrt{43}}{6}
Multiply 2 times 3.
x=\frac{2\sqrt{43}+10}{6}
Now solve the equation x=\frac{10±2\sqrt{43}}{6} when ± is plus. Add 10 to 2\sqrt{43}.
x=\frac{\sqrt{43}+5}{3}
Divide 10+2\sqrt{43} by 6.
x=\frac{10-2\sqrt{43}}{6}
Now solve the equation x=\frac{10±2\sqrt{43}}{6} when ± is minus. Subtract 2\sqrt{43} from 10.
x=\frac{5-\sqrt{43}}{3}
Divide 10-2\sqrt{43} by 6.
x=\frac{\sqrt{43}+5}{3} x=\frac{5-\sqrt{43}}{3}
The equation is now solved.
3x^{2}-10x-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}-10x-6-\left(-6\right)=-\left(-6\right)
Add 6 to both sides of the equation.
3x^{2}-10x=-\left(-6\right)
Subtracting -6 from itself leaves 0.
3x^{2}-10x=6
Subtract -6 from 0.
\frac{3x^{2}-10x}{3}=\frac{6}{3}
Divide both sides by 3.
x^{2}-\frac{10}{3}x=\frac{6}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{10}{3}x=2
Divide 6 by 3.
x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=2+\left(-\frac{5}{3}\right)^{2}
Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{10}{3}x+\frac{25}{9}=2+\frac{25}{9}
Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{43}{9}
Add 2 to \frac{25}{9}.
\left(x-\frac{5}{3}\right)^{2}=\frac{43}{9}
Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.
\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{43}{9}}
Take the square root of both sides of the equation.
x-\frac{5}{3}=\frac{\sqrt{43}}{3} x-\frac{5}{3}=-\frac{\sqrt{43}}{3}
Simplify.
x=\frac{\sqrt{43}+5}{3} x=\frac{5-\sqrt{43}}{3}
Add \frac{5}{3} to both sides of the equation.
x ^ 2 -\frac{10}{3}x -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 = \frac{10}{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{5}{3} - u s = \frac{5}{3} + u
Two numbers r and s sum up to \frac{10}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{10}{3} = \frac{5}{3}. 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{5}{3} - u) (\frac{5}{3} + u) = -2
To solve for unknown quantity u, substitute these in the product equation rs = -2
\frac{25}{9} - u^2 = -2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -2-\frac{25}{9} = -\frac{43}{9}
Simplify the expression by subtracting \frac{25}{9} on both sides
u^2 = \frac{43}{9} u = \pm\sqrt{\frac{43}{9}} = \pm \frac{\sqrt{43}}{3}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{5}{3} - \frac{\sqrt{43}}{3} = -0.519 s = \frac{5}{3} + \frac{\sqrt{43}}{3} = 3.852
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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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
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