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8x^{2}+48x+27=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{-48±\sqrt{48^{2}-4\times 8\times 27}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 48 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-48±\sqrt{2304-4\times 8\times 27}}{2\times 8}
Square 48.
x=\frac{-48±\sqrt{2304-32\times 27}}{2\times 8}
Multiply -4 times 8.
x=\frac{-48±\sqrt{2304-864}}{2\times 8}
Multiply -32 times 27.
x=\frac{-48±\sqrt{1440}}{2\times 8}
Add 2304 to -864.
x=\frac{-48±12\sqrt{10}}{2\times 8}
Take the square root of 1440.
x=\frac{-48±12\sqrt{10}}{16}
Multiply 2 times 8.
x=\frac{12\sqrt{10}-48}{16}
Now solve the equation x=\frac{-48±12\sqrt{10}}{16} when ± is plus. Add -48 to 12\sqrt{10}.
x=\frac{3\sqrt{10}}{4}-3
Divide -48+12\sqrt{10} by 16.
x=\frac{-12\sqrt{10}-48}{16}
Now solve the equation x=\frac{-48±12\sqrt{10}}{16} when ± is minus. Subtract 12\sqrt{10} from -48.
x=-\frac{3\sqrt{10}}{4}-3
Divide -48-12\sqrt{10} by 16.
x=\frac{3\sqrt{10}}{4}-3 x=-\frac{3\sqrt{10}}{4}-3
The equation is now solved.
8x^{2}+48x+27=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.
8x^{2}+48x+27-27=-27
Subtract 27 from both sides of the equation.
8x^{2}+48x=-27
Subtracting 27 from itself leaves 0.
\frac{8x^{2}+48x}{8}=-\frac{27}{8}
Divide both sides by 8.
x^{2}+\frac{48}{8}x=-\frac{27}{8}
Dividing by 8 undoes the multiplication by 8.
x^{2}+6x=-\frac{27}{8}
Divide 48 by 8.
x^{2}+6x+3^{2}=-\frac{27}{8}+3^{2}
Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+6x+9=-\frac{27}{8}+9
Square 3.
x^{2}+6x+9=\frac{45}{8}
Add -\frac{27}{8} to 9.
\left(x+3\right)^{2}=\frac{45}{8}
Factor x^{2}+6x+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+3\right)^{2}}=\sqrt{\frac{45}{8}}
Take the square root of both sides of the equation.
x+3=\frac{3\sqrt{10}}{4} x+3=-\frac{3\sqrt{10}}{4}
Simplify.
x=\frac{3\sqrt{10}}{4}-3 x=-\frac{3\sqrt{10}}{4}-3
Subtract 3 from both sides of the equation.
x ^ 2 +6x +\frac{27}{8} = 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 8
r + s = -6 rs = \frac{27}{8}
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 = -3 - u s = -3 + u
Two numbers r and s sum up to -6 exactly when the average of the two numbers is \frac{1}{2}*-6 = -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>
(-3 - u) (-3 + u) = \frac{27}{8}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{27}{8}
9 - u^2 = \frac{27}{8}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{27}{8}-9 = -\frac{45}{8}
Simplify the expression by subtracting 9 on both sides
u^2 = \frac{45}{8} u = \pm\sqrt{\frac{45}{8}} = \pm \frac{\sqrt{45}}{\sqrt{8}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-3 - \frac{\sqrt{45}}{\sqrt{8}} = -5.372 s = -3 + \frac{\sqrt{45}}{\sqrt{8}} = -0.628
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.