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5x^{2}-40x-75=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(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 5\left(-75\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -40 for b, and -75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-40\right)±\sqrt{1600-4\times 5\left(-75\right)}}{2\times 5}
Square -40.
x=\frac{-\left(-40\right)±\sqrt{1600-20\left(-75\right)}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-40\right)±\sqrt{1600+1500}}{2\times 5}
Multiply -20 times -75.
x=\frac{-\left(-40\right)±\sqrt{3100}}{2\times 5}
Add 1600 to 1500.
x=\frac{-\left(-40\right)±10\sqrt{31}}{2\times 5}
Take the square root of 3100.
x=\frac{40±10\sqrt{31}}{2\times 5}
The opposite of -40 is 40.
x=\frac{40±10\sqrt{31}}{10}
Multiply 2 times 5.
x=\frac{10\sqrt{31}+40}{10}
Now solve the equation x=\frac{40±10\sqrt{31}}{10} when ± is plus. Add 40 to 10\sqrt{31}.
x=\sqrt{31}+4
Divide 40+10\sqrt{31} by 10.
x=\frac{40-10\sqrt{31}}{10}
Now solve the equation x=\frac{40±10\sqrt{31}}{10} when ± is minus. Subtract 10\sqrt{31} from 40.
x=4-\sqrt{31}
Divide 40-10\sqrt{31} by 10.
x=\sqrt{31}+4 x=4-\sqrt{31}
The equation is now solved.
5x^{2}-40x-75=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}-40x-75-\left(-75\right)=-\left(-75\right)
Add 75 to both sides of the equation.
5x^{2}-40x=-\left(-75\right)
Subtracting -75 from itself leaves 0.
5x^{2}-40x=75
Subtract -75 from 0.
\frac{5x^{2}-40x}{5}=\frac{75}{5}
Divide both sides by 5.
x^{2}+\left(-\frac{40}{5}\right)x=\frac{75}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-8x=\frac{75}{5}
Divide -40 by 5.
x^{2}-8x=15
Divide 75 by 5.
x^{2}-8x+\left(-4\right)^{2}=15+\left(-4\right)^{2}
Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-8x+16=15+16
Square -4.
x^{2}-8x+16=31
Add 15 to 16.
\left(x-4\right)^{2}=31
Factor x^{2}-8x+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-4\right)^{2}}=\sqrt{31}
Take the square root of both sides of the equation.
x-4=\sqrt{31} x-4=-\sqrt{31}
Simplify.
x=\sqrt{31}+4 x=4-\sqrt{31}
Add 4 to both sides of the equation.
x ^ 2 -8x -15 = 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 = 8 rs = -15
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 = 4 - u s = 4 + u
Two numbers r and s sum up to 8 exactly when the average of the two numbers is \frac{1}{2}*8 = 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>
(4 - u) (4 + u) = -15
To solve for unknown quantity u, substitute these in the product equation rs = -15
16 - u^2 = -15
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -15-16 = -31
Simplify the expression by subtracting 16 on both sides
u^2 = 31 u = \pm\sqrt{31} = \pm \sqrt{31}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =4 - \sqrt{31} = -1.568 s = 4 + \sqrt{31} = 9.568
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.