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y^{2}+4y-5=0
Subtract 5 from both sides.
a+b=4 ab=-5
To solve the equation, factor y^{2}+4y-5 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.
a=-1 b=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. The only such pair is the system solution.
\left(y-1\right)\left(y+5\right)
Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.
y=1 y=-5
To find equation solutions, solve y-1=0 and y+5=0.
y^{2}+4y-5=0
Subtract 5 from both sides.
a+b=4 ab=1\left(-5\right)=-5
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as y^{2}+ay+by-5. To find a and b, set up a system to be solved.
a=-1 b=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. The only such pair is the system solution.
\left(y^{2}-y\right)+\left(5y-5\right)
Rewrite y^{2}+4y-5 as \left(y^{2}-y\right)+\left(5y-5\right).
y\left(y-1\right)+5\left(y-1\right)
Factor out y in the first and 5 in the second group.
\left(y-1\right)\left(y+5\right)
Factor out common term y-1 by using distributive property.
y=1 y=-5
To find equation solutions, solve y-1=0 and y+5=0.
y^{2}+4y=5
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.
y^{2}+4y-5=5-5
Subtract 5 from both sides of the equation.
y^{2}+4y-5=0
Subtracting 5 from itself leaves 0.
y=\frac{-4±\sqrt{4^{2}-4\left(-5\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}
Square 4.
y=\frac{-4±\sqrt{16+20}}{2}
Multiply -4 times -5.
y=\frac{-4±\sqrt{36}}{2}
Add 16 to 20.
y=\frac{-4±6}{2}
Take the square root of 36.
y=\frac{2}{2}
Now solve the equation y=\frac{-4±6}{2} when ± is plus. Add -4 to 6.
y=1
Divide 2 by 2.
y=-\frac{10}{2}
Now solve the equation y=\frac{-4±6}{2} when ± is minus. Subtract 6 from -4.
y=-5
Divide -10 by 2.
y=1 y=-5
The equation is now solved.
y^{2}+4y=5
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.
y^{2}+4y+2^{2}=5+2^{2}
Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}+4y+4=5+4
Square 2.
y^{2}+4y+4=9
Add 5 to 4.
\left(y+2\right)^{2}=9
Factor y^{2}+4y+4. 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+2\right)^{2}}=\sqrt{9}
Take the square root of both sides of the equation.
y+2=3 y+2=-3
Simplify.
y=1 y=-5
Subtract 2 from both sides of the equation.