Solution
The equation of two curves are
[tex]y_{1}=y=2 \times (5 x+3)\\\\y_{2}=y=x^2-15[/tex]
Equating the two curves that is y values, to find their point of Intersection
→ 2 × (5 x+3)=x²-15
→10 x +6=x²-15⇒⇒Used distributive property of multiplication with respect to addition
→x² -10 x -15 -6=0
→x² -10 x-21=0
For a Quadratic equation of the type
[tex]\rightarrow ax^2+b x +c=0\\\\x=\frac{-b \pm \sqrt{D}}{2 a}\\\\D=b^2-4ac[/tex]
Using discriminant method to solve the problem
[tex]x=\frac{-(-10)\pm\sqrt{(-10)^2-4 \times 1 \times (-21)}}{2 \times 1}\\\\x=\frac{10 \pm \sqrt{100+84}}{2}\\\\x=\frac{10 \pm \sqrt{184}}{2}\\\\x=5 \pm \sqrt{46}\\\\x_{1}=5+6.78\\\\x_{1}=11.78\\\\x_{2}=5-6.78\\\\x_{2}=-1.78[/tex]
The x coordinate of point of intersection are, 11.78 and -1.78.
