Solve for y
y=-4
y=\frac{5}{8}=0.625
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8y^{2}+27y-20=0
Subtract 20 from both sides.
a+b=27 ab=8\left(-20\right)=-160
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8y^{2}+ay+by-20. To find a and b, set up a system to be solved.
-1,160 -2,80 -4,40 -5,32 -8,20 -10,16
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 -160.
-1+160=159 -2+80=78 -4+40=36 -5+32=27 -8+20=12 -10+16=6
Calculate the sum for each pair.
a=-5 b=32
The solution is the pair that gives sum 27.
\left(8y^{2}-5y\right)+\left(32y-20\right)
Rewrite 8y^{2}+27y-20 as \left(8y^{2}-5y\right)+\left(32y-20\right).
y\left(8y-5\right)+4\left(8y-5\right)
Factor out y in the first and 4 in the second group.
\left(8y-5\right)\left(y+4\right)
Factor out common term 8y-5 by using distributive property.
y=\frac{5}{8} y=-4
To find equation solutions, solve 8y-5=0 and y+4=0.
8y^{2}+27y=20
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.
8y^{2}+27y-20=20-20
Subtract 20 from both sides of the equation.
8y^{2}+27y-20=0
Subtracting 20 from itself leaves 0.
y=\frac{-27±\sqrt{27^{2}-4\times 8\left(-20\right)}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 27 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-27±\sqrt{729-4\times 8\left(-20\right)}}{2\times 8}
Square 27.
y=\frac{-27±\sqrt{729-32\left(-20\right)}}{2\times 8}
Multiply -4 times 8.
y=\frac{-27±\sqrt{729+640}}{2\times 8}
Multiply -32 times -20.
y=\frac{-27±\sqrt{1369}}{2\times 8}
Add 729 to 640.
y=\frac{-27±37}{2\times 8}
Take the square root of 1369.
y=\frac{-27±37}{16}
Multiply 2 times 8.
y=\frac{10}{16}
Now solve the equation y=\frac{-27±37}{16} when ± is plus. Add -27 to 37.
y=\frac{5}{8}
Reduce the fraction \frac{10}{16} to lowest terms by extracting and canceling out 2.
y=-\frac{64}{16}
Now solve the equation y=\frac{-27±37}{16} when ± is minus. Subtract 37 from -27.
y=-4
Divide -64 by 16.
y=\frac{5}{8} y=-4
The equation is now solved.
8y^{2}+27y=20
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.
\frac{8y^{2}+27y}{8}=\frac{20}{8}
Divide both sides by 8.
y^{2}+\frac{27}{8}y=\frac{20}{8}
Dividing by 8 undoes the multiplication by 8.
y^{2}+\frac{27}{8}y=\frac{5}{2}
Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.
y^{2}+\frac{27}{8}y+\left(\frac{27}{16}\right)^{2}=\frac{5}{2}+\left(\frac{27}{16}\right)^{2}
Divide \frac{27}{8}, the coefficient of the x term, by 2 to get \frac{27}{16}. Then add the square of \frac{27}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+\frac{27}{8}y+\frac{729}{256}=\frac{5}{2}+\frac{729}{256}
Square \frac{27}{16} by squaring both the numerator and the denominator of the fraction.
y^{2}+\frac{27}{8}y+\frac{729}{256}=\frac{1369}{256}
Add \frac{5}{2} to \frac{729}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(y+\frac{27}{16}\right)^{2}=\frac{1369}{256}
Factor y^{2}+\frac{27}{8}y+\frac{729}{256}. 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(y+\frac{27}{16}\right)^{2}}=\sqrt{\frac{1369}{256}}
Take the square root of both sides of the equation.
y+\frac{27}{16}=\frac{37}{16} y+\frac{27}{16}=-\frac{37}{16}
Simplify.
y=\frac{5}{8} y=-4
Subtract \frac{27}{16} from both sides of the equation.
Examples
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Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
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Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
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