How do you solve this using completing the square? −16t^{2}+32t=−5-16t^{2} +32t=-5−16t^{2}+32t=−5

How do you solve this using completing the square? $-16{t}^{2}+32t=-5-16{t}^{2}+32t=-5-16{t}^{2}+32t=-5$
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To solve an equation by completing the square, you need to rewrite the expression to have a leading coefficient of 1.
Since the leading coefficient of $-16{t}^{2}+32t=-5$ is -16, you need to divide both sides by −16 to get a leading coefficient of 1. This gives ${t}^{2}-2t=\frac{5}{16}$.
To complete the square for an expression of the form ${x}^{2}+bx$, you need to add ${\left(\frac{b}{2}\right)}^{2}$.
For ${t}^{2}-2t,b=-2$ so you need to add ${\left(\frac{-2}{2}\right)}^{2}=\left(-1{\right)}^{2}=1$.
Adding 1 on both sides of ${t}^{2}-2t=\frac{5}{16}$ then gives ${t}^{2}-2t+1=\left(\frac{5}{16}\right)+1$ which simplifies to ${t}^{2}-2t+1=\frac{21}{16}$.
Now that the left side is a perfect square, you need to factor. Since ${x}^{2}+bx+{\left(\frac{b}{2}\right)}^{2}$ factors to $\left(x+\frac{b}{2}\right)$, then ${t}^{2}-2t+1$ factors to ${\left(t-1\right)}^{2}$ since $\frac{b}{2}=-\frac{2}{2}=-11.$
The factored equation is then ${\left(t-1\right)}^{2}=\frac{21}{16}.$
Square rooting both sides gives $\sqrt{{\left(t-1\right)}^{2}}=±\frac{21}{16}$ which simplifies to $t-1=±\sqrt{\left(\frac{21}{4}\right)}$
Adding 1 on both sides then gives $t=1±\sqrt{\frac{21}{4}}$