Question

100-(z-19)²=0
25-(13+t)²=0

Answer 1

$1.$$-(z-19)^2=-100$
$(z-19)^2=100$
$(z-19)^{2* \frac{1}{2} }=\pm \sqrt{100}$
$(z-19)=\pm \sqrt{100}$
$(z-19)=\pm10$
$z=29;9$

$2.$$-(13+t)^2=-25$
$(13+t)^2=25$
$(13+t)^{2* \frac{1}{2} }=\pm \sqrt{25}$
$(13+t)=\pm \sqrt{25}$
$(13+t)=\pm5$
$t=-8;-18$

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