Найдите корень уравнения sin[latex] \frac{px}{2} [/latex]=[latex]x^{2} [/latex]+6x+10

[latex]sin \frac{ \pi x}{2} =x^2+6x+10 \\ -1\leq sin \frac{ \pi x}{2} \leq1, \\ x^{2}+6x+10=(x^{2}+6x+9)+1=(x+3)^2+1 \geq 1 [/latex] [latex]\left \{ {{sin \frac{ \pi x}{2}=1} \atop {(x+3)^2+1=1}} \right.[/latex], n ∈ Z [latex]\left \{ {{\frac{ \pi x}{2}=\frac{ \pi}{2}+2 \pi n} \atop {(x+3)^2=0}} \right. [/latex], n ∈ Z [latex]\left \{ {{x=1+4n} \atop {x=-3}} \right.[/latex], n ∈ Z x = -3