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10x^{2}-15x+2=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(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 10\times 2}}{2\times 10}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, -15 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-15\right)±\sqrt{225-4\times 10\times 2}}{2\times 10}
Square -15.
x=\frac{-\left(-15\right)±\sqrt{225-40\times 2}}{2\times 10}
Multiply -4 times 10.
x=\frac{-\left(-15\right)±\sqrt{225-80}}{2\times 10}
Multiply -40 times 2.
x=\frac{-\left(-15\right)±\sqrt{145}}{2\times 10}
Add 225 to -80.
x=\frac{15±\sqrt{145}}{2\times 10}
The opposite of -15 is 15.
x=\frac{15±\sqrt{145}}{20}
Multiply 2 times 10.
x=\frac{\sqrt{145}+15}{20}
Now solve the equation x=\frac{15±\sqrt{145}}{20} when ± is plus. Add 15 to \sqrt{145}.
x=\frac{\sqrt{145}}{20}+\frac{3}{4}
Divide 15+\sqrt{145} by 20.
x=\frac{15-\sqrt{145}}{20}
Now solve the equation x=\frac{15±\sqrt{145}}{20} when ± is minus. Subtract \sqrt{145} from 15.
x=-\frac{\sqrt{145}}{20}+\frac{3}{4}
Divide 15-\sqrt{145} by 20.
x=\frac{\sqrt{145}}{20}+\frac{3}{4} x=-\frac{\sqrt{145}}{20}+\frac{3}{4}
The equation is now solved.
10x^{2}-15x+2=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.
10x^{2}-15x+2-2=-2
Subtract 2 from both sides of the equation.
10x^{2}-15x=-2
Subtracting 2 from itself leaves 0.
\frac{10x^{2}-15x}{10}=-\frac{2}{10}
Divide both sides by 10.
x^{2}+\left(-\frac{15}{10}\right)x=-\frac{2}{10}
Dividing by 10 undoes the multiplication by 10.
x^{2}-\frac{3}{2}x=-\frac{2}{10}
Reduce the fraction \frac{-15}{10} to lowest terms by extracting and canceling out 5.
x^{2}-\frac{3}{2}x=-\frac{1}{5}
Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=-\frac{1}{5}+\left(-\frac{3}{4}\right)^{2}
Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{1}{5}+\frac{9}{16}
Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{29}{80}
Add -\frac{1}{5} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{4}\right)^{2}=\frac{29}{80}
Factor x^{2}-\frac{3}{2}x+\frac{9}{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-\frac{3}{4}\right)^{2}}=\sqrt{\frac{29}{80}}
Take the square root of both sides of the equation.
x-\frac{3}{4}=\frac{\sqrt{145}}{20} x-\frac{3}{4}=-\frac{\sqrt{145}}{20}
Simplify.
x=\frac{\sqrt{145}}{20}+\frac{3}{4} x=-\frac{\sqrt{145}}{20}+\frac{3}{4}
Add \frac{3}{4} to both sides of the equation.
x ^ 2 -\frac{3}{2}x +\frac{1}{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 10
r + s = \frac{3}{2} rs = \frac{1}{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{3}{4} - u s = \frac{3}{4} + u
Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{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>
(\frac{3}{4} - u) (\frac{3}{4} + u) = \frac{1}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{5}
\frac{9}{16} - u^2 = \frac{1}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{5}-\frac{9}{16} = -\frac{29}{80}
Simplify the expression by subtracting \frac{9}{16} on both sides
u^2 = \frac{29}{80} u = \pm\sqrt{\frac{29}{80}} = \pm \frac{\sqrt{29}}{\sqrt{80}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{4} - \frac{\sqrt{29}}{\sqrt{80}} = 0.148 s = \frac{3}{4} + \frac{\sqrt{29}}{\sqrt{80}} = 1.352
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.