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-3.75=x^{2}+2x+1-4
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.
-3.75=x^{2}+2x-3
Subtract 4 from 1 to get -3.
x^{2}+2x-3=-3.75
Swap sides so that all variable terms are on the left hand side.
x^{2}+2x-3+3.75=0
Add 3.75 to both sides.
x^{2}+2x+0.75=0
Add -3 and 3.75 to get 0.75.
x=\frac{-2±\sqrt{2^{2}-4\times 0.75}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and 0.75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2±\sqrt{4-4\times 0.75}}{2}
Square 2.
x=\frac{-2±\sqrt{4-3}}{2}
Multiply -4 times 0.75.
x=\frac{-2±\sqrt{1}}{2}
Add 4 to -3.
x=\frac{-2±1}{2}
Take the square root of 1.
x=-\frac{1}{2}
Now solve the equation x=\frac{-2±1}{2} when ± is plus. Add -2 to 1.
x=-\frac{3}{2}
Now solve the equation x=\frac{-2±1}{2} when ± is minus. Subtract 1 from -2.
x=-\frac{1}{2} x=-\frac{3}{2}
The equation is now solved.
-3.75=x^{2}+2x+1-4
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.
-3.75=x^{2}+2x-3
Subtract 4 from 1 to get -3.
x^{2}+2x-3=-3.75
Swap sides so that all variable terms are on the left hand side.
x^{2}+2x=-3.75+3
Add 3 to both sides.
x^{2}+2x=-0.75
Add -3.75 and 3 to get -0.75.
x^{2}+2x+1^{2}=-0.75+1^{2}
Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+2x+1=-0.75+1
Square 1.
x^{2}+2x+1=0.25
Add -0.75 to 1.
\left(x+1\right)^{2}=0.25
Factor x^{2}+2x+1. 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+1\right)^{2}}=\sqrt{0.25}
Take the square root of both sides of the equation.
x+1=\frac{1}{2} x+1=-\frac{1}{2}
Simplify.
x=-\frac{1}{2} x=-\frac{3}{2}
Subtract 1 from both sides of the equation.