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-3x^{2}+12x-5=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{-12±\sqrt{12^{2}-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 12 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-12±\sqrt{144-4\left(-3\right)\left(-5\right)}}{2\left(-3\right)}
Square 12.
x=\frac{-12±\sqrt{144+12\left(-5\right)}}{2\left(-3\right)}
Multiply -4 times -3.
x=\frac{-12±\sqrt{144-60}}{2\left(-3\right)}
Multiply 12 times -5.
x=\frac{-12±\sqrt{84}}{2\left(-3\right)}
Add 144 to -60.
x=\frac{-12±2\sqrt{21}}{2\left(-3\right)}
Take the square root of 84.
x=\frac{-12±2\sqrt{21}}{-6}
Multiply 2 times -3.
x=\frac{2\sqrt{21}-12}{-6}
Now solve the equation x=\frac{-12±2\sqrt{21}}{-6} when ± is plus. Add -12 to 2\sqrt{21}.
x=-\frac{\sqrt{21}}{3}+2
Divide -12+2\sqrt{21} by -6.
x=\frac{-2\sqrt{21}-12}{-6}
Now solve the equation x=\frac{-12±2\sqrt{21}}{-6} when ± is minus. Subtract 2\sqrt{21} from -12.
x=\frac{\sqrt{21}}{3}+2
Divide -12-2\sqrt{21} by -6.
x=-\frac{\sqrt{21}}{3}+2 x=\frac{\sqrt{21}}{3}+2
The equation is now solved.
-3x^{2}+12x-5=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}+12x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
-3x^{2}+12x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
-3x^{2}+12x=5
Subtract -5 from 0.
\frac{-3x^{2}+12x}{-3}=\frac{5}{-3}
Divide both sides by -3.
x^{2}+\frac{12}{-3}x=\frac{5}{-3}
Dividing by -3 undoes the multiplication by -3.
x^{2}-4x=\frac{5}{-3}
Divide 12 by -3.
x^{2}-4x=-\frac{5}{3}
Divide 5 by -3.
x^{2}-4x+\left(-2\right)^{2}=-\frac{5}{3}+\left(-2\right)^{2}
Divide -4, the coefficient of the x term, by 2 to get -2. Then add the square of -2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-4x+4=-\frac{5}{3}+4
Square -2.
x^{2}-4x+4=\frac{7}{3}
Add -\frac{5}{3} to 4.
\left(x-2\right)^{2}=\frac{7}{3}
Factor x^{2}-4x+4. 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-2\right)^{2}}=\sqrt{\frac{7}{3}}
Take the square root of both sides of the equation.
x-2=\frac{\sqrt{21}}{3} x-2=-\frac{\sqrt{21}}{3}
Simplify.
x=\frac{\sqrt{21}}{3}+2 x=-\frac{\sqrt{21}}{3}+2
Add 2 to both sides of the equation.
x ^ 2 -4x +\frac{5}{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.
r + s = 4 rs = \frac{5}{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 = 2 - u s = 2 + u
Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 2. 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>
(2 - u) (2 + u) = \frac{5}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{5}{3}
4 - u^2 = \frac{5}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{5}{3}-4 = -\frac{7}{3}
Simplify the expression by subtracting 4 on both sides
u^2 = \frac{7}{3} u = \pm\sqrt{\frac{7}{3}} = \pm \frac{\sqrt{7}}{\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 =2 - \frac{\sqrt{7}}{\sqrt{3}} = 0.472 s = 2 + \frac{\sqrt{7}}{\sqrt{3}} = 3.528
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.