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5x^{2}-7x-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{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -7 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-7\right)±\sqrt{49-4\times 5\left(-3\right)}}{2\times 5}
Square -7.
x=\frac{-\left(-7\right)±\sqrt{49-20\left(-3\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-7\right)±\sqrt{49+60}}{2\times 5}
Multiply -20 times -3.
x=\frac{-\left(-7\right)±\sqrt{109}}{2\times 5}
Add 49 to 60.
x=\frac{7±\sqrt{109}}{2\times 5}
The opposite of -7 is 7.
x=\frac{7±\sqrt{109}}{10}
Multiply 2 times 5.
x=\frac{\sqrt{109}+7}{10}
Now solve the equation x=\frac{7±\sqrt{109}}{10} when ± is plus. Add 7 to \sqrt{109}.
x=\frac{7-\sqrt{109}}{10}
Now solve the equation x=\frac{7±\sqrt{109}}{10} when ± is minus. Subtract \sqrt{109} from 7.
x=\frac{\sqrt{109}+7}{10} x=\frac{7-\sqrt{109}}{10}
The equation is now solved.
5x^{2}-7x-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.
5x^{2}-7x-3-\left(-3\right)=-\left(-3\right)
Add 3 to both sides of the equation.
5x^{2}-7x=-\left(-3\right)
Subtracting -3 from itself leaves 0.
5x^{2}-7x=3
Subtract -3 from 0.
\frac{5x^{2}-7x}{5}=\frac{3}{5}
Divide both sides by 5.
x^{2}-\frac{7}{5}x=\frac{3}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{7}{5}x+\left(-\frac{7}{10}\right)^{2}=\frac{3}{5}+\left(-\frac{7}{10}\right)^{2}
Divide -\frac{7}{5}, the coefficient of the x term, by 2 to get -\frac{7}{10}. Then add the square of -\frac{7}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{3}{5}+\frac{49}{100}
Square -\frac{7}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{7}{5}x+\frac{49}{100}=\frac{109}{100}
Add \frac{3}{5} to \frac{49}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{7}{10}\right)^{2}=\frac{109}{100}
Factor x^{2}-\frac{7}{5}x+\frac{49}{100}. 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{7}{10}\right)^{2}}=\sqrt{\frac{109}{100}}
Take the square root of both sides of the equation.
x-\frac{7}{10}=\frac{\sqrt{109}}{10} x-\frac{7}{10}=-\frac{\sqrt{109}}{10}
Simplify.
x=\frac{\sqrt{109}+7}{10} x=\frac{7-\sqrt{109}}{10}
Add \frac{7}{10} to both sides of the equation.
x ^ 2 -\frac{7}{5}x -\frac{3}{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 5
r + s = \frac{7}{5} rs = -\frac{3}{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{7}{10} - u s = \frac{7}{10} + u
Two numbers r and s sum up to \frac{7}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{7}{5} = \frac{7}{10}. 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{7}{10} - u) (\frac{7}{10} + u) = -\frac{3}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{5}
\frac{49}{100} - u^2 = -\frac{3}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{3}{5}-\frac{49}{100} = -\frac{109}{100}
Simplify the expression by subtracting \frac{49}{100} on both sides
u^2 = \frac{109}{100} u = \pm\sqrt{\frac{109}{100}} = \pm \frac{\sqrt{109}}{10}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{7}{10} - \frac{\sqrt{109}}{10} = -0.344 s = \frac{7}{10} + \frac{\sqrt{109}}{10} = 1.744
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.