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10x^{2}+3x-3=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{-3±\sqrt{3^{2}-4\times 10\left(-3\right)}}{2\times 10}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 3 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\times 10\left(-3\right)}}{2\times 10}
Square 3.
x=\frac{-3±\sqrt{9-40\left(-3\right)}}{2\times 10}
Multiply -4 times 10.
x=\frac{-3±\sqrt{9+120}}{2\times 10}
Multiply -40 times -3.
x=\frac{-3±\sqrt{129}}{2\times 10}
Add 9 to 120.
x=\frac{-3±\sqrt{129}}{20}
Multiply 2 times 10.
x=\frac{\sqrt{129}-3}{20}
Now solve the equation x=\frac{-3±\sqrt{129}}{20} when ± is plus. Add -3 to \sqrt{129}.
x=\frac{-\sqrt{129}-3}{20}
Now solve the equation x=\frac{-3±\sqrt{129}}{20} when ± is minus. Subtract \sqrt{129} from -3.
x=\frac{\sqrt{129}-3}{20} x=\frac{-\sqrt{129}-3}{20}
The equation is now solved.
10x^{2}+3x-3=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}+3x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
10x^{2}+3x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
10x^{2}+3x=3
Subtract -3 from 0.
\frac{10x^{2}+3x}{10}=\frac{3}{10}
Divide both sides by 10.
x^{2}+\frac{3}{10}x=\frac{3}{10}
Dividing by 10 undoes the multiplication by 10.
x^{2}+\frac{3}{10}x+\left(\frac{3}{20}\right)^{2}=\frac{3}{10}+\left(\frac{3}{20}\right)^{2}
Divide \frac{3}{10}, the coefficient of the x term, by 2 to get \frac{3}{20}. Then add the square of \frac{3}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{3}{10}x+\frac{9}{400}=\frac{3}{10}+\frac{9}{400}
Square \frac{3}{20} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{3}{10}x+\frac{9}{400}=\frac{129}{400}
Add \frac{3}{10} to \frac{9}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{3}{20}\right)^{2}=\frac{129}{400}
Factor x^{2}+\frac{3}{10}x+\frac{9}{400}. 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}{20}\right)^{2}}=\sqrt{\frac{129}{400}}
Take the square root of both sides of the equation.
x+\frac{3}{20}=\frac{\sqrt{129}}{20} x+\frac{3}{20}=-\frac{\sqrt{129}}{20}
Simplify.
x=\frac{\sqrt{129}-3}{20} x=\frac{-\sqrt{129}-3}{20}
Subtract \frac{3}{20} from both sides of the equation.
x ^ 2 +\frac{3}{10}x -\frac{3}{10} = 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}{10} rs = -\frac{3}{10}
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}{20} - u s = -\frac{3}{20} + u
Two numbers r and s sum up to -\frac{3}{10} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{10} = -\frac{3}{20}. 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}{20} - u) (-\frac{3}{20} + u) = -\frac{3}{10}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{10}
\frac{9}{400} - u^2 = -\frac{3}{10}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{10}-\frac{9}{400} = -\frac{129}{400}
Simplify the expression by subtracting \frac{9}{400} on both sides
u^2 = \frac{129}{400} u = \pm\sqrt{\frac{129}{400}} = \pm \frac{\sqrt{129}}{20}
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}{20} - \frac{\sqrt{129}}{20} = -0.718 s = -\frac{3}{20} + \frac{\sqrt{129}}{20} = 0.418
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.