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4x^2+6x-5=35
We move all terms to the left:
4x^2+6x-5-(35)=0
We add all the numbers together, and all the variables
4x^2+6x-40=0
a = 4; b = 6; c = -40;
Δ = b2-4ac
Δ = 62-4·4·(-40)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{676}=26$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-26}{2*4}=\frac{-32}{8} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+26}{2*4}=\frac{20}{8} =2+1/2 $
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