Resposta :
a) 4t²-25=0
4t²=25
t²=25
4
t=[tex] \sqrt{25/4} [/tex]
t=[tex] \frac{5}{2} [/tex]
b)2x²-14x+24=0
Δ = b² – 4 . a . c
Δ =14²-4x2x24
Δ =196-192=4
x = – b ± √Δ ÷ 2a
x=14±[tex] \sqrt{4 } [/tex]
2x2
x=14±2
4
x1=14+2= 16=4
4 4
x2=14-2=12=3
4 4
4t²=25
t²=25
4
t=[tex] \sqrt{25/4} [/tex]
t=[tex] \frac{5}{2} [/tex]
b)2x²-14x+24=0
Δ = b² – 4 . a . c
Δ =14²-4x2x24
Δ =196-192=4
x = – b ± √Δ ÷ 2a
x=14±[tex] \sqrt{4 } [/tex]
2x2
x=14±2
4
x1=14+2= 16=4
4 4
x2=14-2=12=3
4 4