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5x^{2}+x-4=0
Divide both sides by 2.
a+b=1 ab=5\left(-4\right)=-20
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-4. To find a and b, set up a system to be solved.
-1,20 -2,10 -4,5
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -20.
-1+20=19 -2+10=8 -4+5=1
Calculate the sum for each pair.
a=-4 b=5
The solution is the pair that gives sum 1.
\left(5x^{2}-4x\right)+\left(5x-4\right)
Rewrite 5x^{2}+x-4 as \left(5x^{2}-4x\right)+\left(5x-4\right).
x\left(5x-4\right)+5x-4
Factor out x in 5x^{2}-4x.
\left(5x-4\right)\left(x+1\right)
Factor out common term 5x-4 by using distributive property.
x=\frac{4}{5} x=-1
To find equation solutions, solve 5x-4=0 and x+1=0.
10x^{2}+2x-8=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{-2±\sqrt{2^{2}-4\times 10\left(-8\right)}}{2\times 10}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 10\left(-8\right)}}{2\times 10}
Square 2.
x=\frac{-2±\sqrt{4-40\left(-8\right)}}{2\times 10}
Multiply -4 times 10.
x=\frac{-2±\sqrt{4+320}}{2\times 10}
Multiply -40 times -8.
x=\frac{-2±\sqrt{324}}{2\times 10}
Add 4 to 320.
x=\frac{-2±18}{2\times 10}
Take the square root of 324.
x=\frac{-2±18}{20}
Multiply 2 times 10.
x=\frac{16}{20}
Now solve the equation x=\frac{-2±18}{20} when ± is plus. Add -2 to 18.
x=\frac{4}{5}
Reduce the fraction \frac{16}{20} to lowest terms by extracting and canceling out 4.
x=-\frac{20}{20}
Now solve the equation x=\frac{-2±18}{20} when ± is minus. Subtract 18 from -2.
x=-1
Divide -20 by 20.
x=\frac{4}{5} x=-1
The equation is now solved.
10x^{2}+2x-8=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}+2x-8-\left(-8\right)=-\left(-8\right)
Add 8 to both sides of the equation.
10x^{2}+2x=-\left(-8\right)
Subtracting -8 from itself leaves 0.
10x^{2}+2x=8
Subtract -8 from 0.
\frac{10x^{2}+2x}{10}=\frac{8}{10}
Divide both sides by 10.
x^{2}+\frac{2}{10}x=\frac{8}{10}
Dividing by 10 undoes the multiplication by 10.
x^{2}+\frac{1}{5}x=\frac{8}{10}
Reduce the fraction \frac{2}{10} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{1}{5}x=\frac{4}{5}
Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{1}{5}x+\left(\frac{1}{10}\right)^{2}=\frac{4}{5}+\left(\frac{1}{10}\right)^{2}
Divide \frac{1}{5}, the coefficient of the x term, by 2 to get \frac{1}{10}. Then add the square of \frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{4}{5}+\frac{1}{100}
Square \frac{1}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{81}{100}
Add \frac{4}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{1}{10}\right)^{2}=\frac{81}{100}
Factor x^{2}+\frac{1}{5}x+\frac{1}{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{1}{10}\right)^{2}}=\sqrt{\frac{81}{100}}
Take the square root of both sides of the equation.
x+\frac{1}{10}=\frac{9}{10} x+\frac{1}{10}=-\frac{9}{10}
Simplify.
x=\frac{4}{5} x=-1
Subtract \frac{1}{10} from both sides of the equation.
x ^ 2 +\frac{1}{5}x -\frac{4}{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{1}{5} rs = -\frac{4}{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{1}{10} - u s = -\frac{1}{10} + u
Two numbers r and s sum up to -\frac{1}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{5} = -\frac{1}{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{1}{10} - u) (-\frac{1}{10} + u) = -\frac{4}{5}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{4}{5}
\frac{1}{100} - u^2 = -\frac{4}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{4}{5}-\frac{1}{100} = -\frac{81}{100}
Simplify the expression by subtracting \frac{1}{100} on both sides
u^2 = \frac{81}{100} u = \pm\sqrt{\frac{81}{100}} = \pm \frac{9}{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{1}{10} - \frac{9}{10} = -1 s = -\frac{1}{10} + \frac{9}{10} = 0.800
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.