a) We have the equation
[tex]\begin{gathered} x^2=49 \\ x=\sqrt[]{49} \\ x=\pm7\text{ (two real solutions: -7 and +7)} \end{gathered}[/tex]
b) This equation is similar to a), and have 2 real solutions.
[tex]\begin{gathered} x^2-74=0 \\ x^2=74 \\ x=\sqrt[]{74} \\ x\approx\pm8.6 \end{gathered}[/tex]
c) This equation can be factorized as:
[tex]\begin{gathered} x^2-10x+25 \\ x^2-2\cdot5x+5^2 \\ (x-5)^2 \end{gathered}[/tex]
It has one real solution (x=5).
d)
[tex]\begin{gathered} 3x^2-6x=29 \\ 3x^2-6x-29=0 \\ x=\frac{6\pm\sqrt[]{36-4\cdot3\cdot(-29)}}{2\cdot3}\text{ \lbrack{}applying the quadratic equation for the roots\rbrack} \\ x=\frac{6}{6}\pm\frac{\sqrt[]{36+348}}{6} \\ x=1\pm\frac{\sqrt[]{384}}{6}\longrightarrow\text{ Two real solutions.} \end{gathered}[/tex]
e)
[tex]\begin{gathered} 2x^2-6x+10=0 \\ 2(x^2-3x+5)=0 \\ x^2-3x+5=0 \\ x=\frac{-(-3)\pm\sqrt[]{(-3)^2-4\cdot1\cdot5}}{2\cdot1} \\ x=\frac{3}{2}\pm\frac{\sqrt[]{9-20}}{2} \\ x=\frac{3}{2}\pm\frac{\sqrt[]{-11}}{2} \\ x=\frac{3}{2}\pm\frac{\sqrt[]{11}}{2}\cdot\sqrt[]{-1} \\ x=\frac{3}{2}\pm\frac{\sqrt[]{11}}{2}i\longrightarrow\text{ Two complex solutions (not real)} \end{gathered}[/tex]
