Solve for y
y=-2
y=2
y=1
y=-1
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t^{2}-5t+4=0
Substitute t for y^{2}.
t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 4}}{2}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -5 for b, and 4 for c in the quadratic formula.
t=\frac{5±3}{2}
Do the calculations.
t=4 t=1
Solve the equation t=\frac{5±3}{2} when ± is plus and when ± is minus.
y=2 y=-2 y=1 y=-1
Since y=t^{2}, the solutions are obtained by evaluating y=±\sqrt{t} for each t.
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