Quadratic equation question

Question

7(t − 1)² − 29(t − 1) = −4

x⁻² − x⁻¹ − 30 = 0

Answer 1

let t-1=z

have then

7z²-29z+4=0

use quadratic formula to find valid z roots

z=(29/14)+/-(√29²-4*4*7)/14

z=29/14+/-27/14

(t-1)=1/7,,,t=8/7

or

t-1=4,,,t=5

for second problem, multiply both sides of equation by -x²

have 30x²+x-1=0

use formula to find valid x

x=(-1/60)+/-(√(1²+4*30)/60)

x=-12/60=-1/5

or

x=10/60= 1/6

Answer 2

7(t − 1)² − 29(t − 1) = −4

7(t − 1)² − 29(t − 1) + 4 = 0

(7(t - 1) - 1)((t - 1) - 4)) = 0

7(t - 1) = 1

or

(t - 1) = 4

x⁻² − x⁻¹ − 30 = 0

(x^-1 - 6)(x^-1 + 5) = 0

You solve for x