Solve for x
x=\frac{2\sqrt{15}}{3}+4\approx 6.581988897
x=-\frac{2\sqrt{15}}{3}+4\approx 1.418011103
Graph
Share
Copied to clipboard
3x^{2}-24x+28=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(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 3\times 28}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -24 for b, and 28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-24\right)±\sqrt{576-4\times 3\times 28}}{2\times 3}
Square -24.
x=\frac{-\left(-24\right)±\sqrt{576-12\times 28}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-24\right)±\sqrt{576-336}}{2\times 3}
Multiply -12 times 28.
x=\frac{-\left(-24\right)±\sqrt{240}}{2\times 3}
Add 576 to -336.
x=\frac{-\left(-24\right)±4\sqrt{15}}{2\times 3}
Take the square root of 240.
x=\frac{24±4\sqrt{15}}{2\times 3}
The opposite of -24 is 24.
x=\frac{24±4\sqrt{15}}{6}
Multiply 2 times 3.
x=\frac{4\sqrt{15}+24}{6}
Now solve the equation x=\frac{24±4\sqrt{15}}{6} when ± is plus. Add 24 to 4\sqrt{15}.
x=\frac{2\sqrt{15}}{3}+4
Divide 24+4\sqrt{15} by 6.
x=\frac{24-4\sqrt{15}}{6}
Now solve the equation x=\frac{24±4\sqrt{15}}{6} when ± is minus. Subtract 4\sqrt{15} from 24.
x=-\frac{2\sqrt{15}}{3}+4
Divide 24-4\sqrt{15} by 6.
x=\frac{2\sqrt{15}}{3}+4 x=-\frac{2\sqrt{15}}{3}+4
The equation is now solved.
3x^{2}-24x+28=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}-24x+28-28=-28
Subtract 28 from both sides of the equation.
3x^{2}-24x=-28
Subtracting 28 from itself leaves 0.
\frac{3x^{2}-24x}{3}=-\frac{28}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{24}{3}\right)x=-\frac{28}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-8x=-\frac{28}{3}
Divide -24 by 3.
x^{2}-8x+\left(-4\right)^{2}=-\frac{28}{3}+\left(-4\right)^{2}
Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-8x+16=-\frac{28}{3}+16
Square -4.
x^{2}-8x+16=\frac{20}{3}
Add -\frac{28}{3} to 16.
\left(x-4\right)^{2}=\frac{20}{3}
Factor x^{2}-8x+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-4\right)^{2}}=\sqrt{\frac{20}{3}}
Take the square root of both sides of the equation.
x-4=\frac{2\sqrt{15}}{3} x-4=-\frac{2\sqrt{15}}{3}
Simplify.
x=\frac{2\sqrt{15}}{3}+4 x=-\frac{2\sqrt{15}}{3}+4
Add 4 to both sides of the equation.
x ^ 2 -8x +\frac{28}{3} = 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 = 8 rs = \frac{28}{3}
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 = 4 - u s = 4 + u
Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 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>
(4 - u) (4 + u) = \frac{28}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{28}{3}
16 - u^2 = \frac{28}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{28}{3}-16 = -\frac{20}{3}
Simplify the expression by subtracting 16 on both sides
u^2 = \frac{20}{3} u = \pm\sqrt{\frac{20}{3}} = \pm \frac{\sqrt{20}}{\sqrt{3}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =4 - \frac{\sqrt{20}}{\sqrt{3}} = 1.418 s = 4 + \frac{\sqrt{20}}{\sqrt{3}} = 6.582
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}