Question

5t⁴+12t+7
Find zeros and verify

Answer 1

$let \: {t}^{2} = x \\ t = \sqrt{x} \\ then5 {x}^{2} 12 \sqrt{x} + 7 = 0 \\ x = - b + \sqrt{{b}^{2} - 4ac} \div 2a \\ where \\ a = 5 \\ b = 12 \\ c = 7 \\ we \: get \: \:$