Answer: [tex]y_1=-\frac{1}{2}\\\\y_2=-\frac{3}{2}[/tex]Step-by-step explanation: The Quadratic formula is: [tex]y=\frac{-b\±\sqrt{b^2-4ac} }{2a}[/tex] Given the equation [tex]4y^2+ 8y +7 =4[/tex], you need to subtract 4 from both sides: [tex]4y^2+ 8y +7 -4=4-4[/tex] [tex]4y^2+ 8y +3 =0[/tex] Now you can identify that: [tex]a=4\\b=8\\c=3[/tex] Then you can substitute these values into the Quadratic formula. Therefore, you get these solutions: [tex]y=\frac{-8\±\sqrt{8^2-4(4)(3)} }{2(4)}[/tex] [tex]y_1=-\frac{1}{2}\\\\y_2=-\frac{3}{2}[/tex]