Question

решите уравнение
-7z^2+(3z-4)^2-2(4+z)(z-4)=0

Answer 1

$- 7 {z}^{2} + (3z - 4 {)}^{2} - 2(4 + z)(z - 4) = 0 \\ - 7 {z}^{2} + 9 {z}^{2} - 24z + 16 - 2( {z}^{2} - 16) = 0 \\ - 7 {z}^{2} + 9 {z}^{2} - 24z + 16 - 2 {z}^{2} + 32 = 0 \\ 0 - 24z + 48 = 0 \\ - 24z + 48 = 0 \\ - 24z = - 48 \\ z = 2$

Answer 2

-7z²+(3z-4)²-2(4+z)(z-4) =0

-7z²+9z²-24z+16-2(z+4)(z-4)=0

-7z²+9z²-24z+16-2(z²-16)=0

-7z²+9z²-24z+16-2z²+32=0

0-24z+48=0

-24z+48=0

-24z=0-48

-24z=-48

Z=48:24

Z=2

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Удачи)))