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5x^2=4x+7
We move all terms to the left:
5x^2-(4x+7)=0
We get rid of parentheses
5x^2-4x-7=0
a = 5; b = -4; c = -7;
Δ = b2-4ac
Δ = -42-4·5·(-7)
Δ = 156
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{39}}{2*5}=\frac{4-2\sqrt{39}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{39}}{2*5}=\frac{4+2\sqrt{39}}{10} $
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